Merge branch 'master' of github.com:youngyangyang04/leetcode-master

problems/0001.两数之和.md

@@ -335,6 +335,20 @@ public class Solution {
 }
 ```
 
+Dart:
+```dart
+List<int> twoSum(List<int> nums, int target) {  
+    var tmp = [];
+    for (var i = 0; i < nums.length; i++) {
+        var rest = target - nums[i];
+        if(tmp.contains(rest)){
+            return [tmp.indexOf(rest), i];
+        }
+        tmp.add(nums[i]);
+    }
+    return [0 , 0];
+}
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0015.三数之和.md

@@ -95,7 +95,7 @@ public:
 
 拿这个nums数组来举例，首先将数组排序，然后有一层for循环，i从下标0的地方开始，同时定一个下标left 定义在i+1的位置上，定义下标right 在数组结尾的位置上。
 
-依然还是在数组中找到 abc 使得a + b +c =0，我们这里相当于  a = nums[i] b = nums[left]  c = nums[right]。
+依然还是在数组中找到 abc 使得a + b +c =0，我们这里相当于  a = nums[i]，b = nums[left]，c = nums[right]。
 
 接下来如何移动left 和right呢， 如果nums[i] + nums[left] + nums[right] > 0  就说明 此时三数之和大了，因为数组是排序后了，所以right下标就应该向左移动，这样才能让三数之和小一些。
 
@@ -411,6 +411,76 @@ var threeSum = function(nums) {
     return res
 };
 ```
+
+解法二：nSum通用解法。递归
+
+```js
+/**
+ *  nsum通用解法，支持2sum，3sum，4sum...等等
+ *  时间复杂度分析：
+ *  1. n = 2时，时间复杂度O(NlogN)，排序所消耗的时间。、
+ *  2. n > 2时，时间复杂度为O(N^n-1)，即N的n-1次方，至少是2次方，此时可省略排序所消耗的时间。举例：3sum为O(n^2)，4sum为O(n^3)
+ * @param {number[]} nums
+ * @return {number[][]}
+ */
+var threeSum = function (nums) {
+    // nsum通用解法核心方法
+    function nSumTarget(nums, n, start, target) {
+        // 前提：nums要先排序好
+        let res = [];
+        if (n === 2) {
+            res = towSumTarget(nums, start, target);
+        } else {
+            for (let i = start; i < nums.length; i++) {
+                // 递归求(n - 1)sum
+                let subRes = nSumTarget(
+                    nums,
+                    n - 1,
+                    i + 1,
+                    target - nums[i]
+                );
+                for (let j = 0; j < subRes.length; j++) {
+                    res.push([nums[i], ...subRes[j]]);
+                }
+                // 跳过相同元素
+                while (nums[i] === nums[i + 1]) i++;
+            }
+        }
+        return res;
+    }
+
+    function towSumTarget(nums, start, target) {
+        // 前提：nums要先排序好
+        let res = [];
+        let len = nums.length;
+        let left = start;
+        let right = len - 1;
+        while (left < right) {
+            let sum = nums[left] + nums[right];
+            if (sum < target) {
+                while (nums[left] === nums[left + 1]) left++;
+                left++;
+            } else if (sum > target) {
+                while (nums[right] === nums[right - 1]) right--;
+                right--;
+            } else {
+                // 相等
+                res.push([nums[left], nums[right]]);
+                // 跳过相同元素
+                while (nums[left] === nums[left + 1]) left++;
+                while (nums[right] === nums[right - 1]) right--;
+                left++;
+                right--;
+            }
+        }
+        return res;
+    }
+    nums.sort((a, b) => a - b);
+    // n = 3，此时求3sum之和
+    return nSumTarget(nums, 3, 0, 0);
+};
+```
+
 TypeScript:
 
 ```typescript
@@ -550,6 +620,71 @@ func threeSum(_ nums: [Int]) -> [[Int]] {
 }
 ```
 
+Rust:
+```Rust
+// 哈希解法
+use std::collections::HashSet;
+impl Solution {
+    pub fn three_sum(nums: Vec<i32>) -> Vec<Vec<i32>> {
+        let mut result: Vec<Vec<i32>> = Vec::new();
+        let mut nums = nums;
+        nums.sort();
+        let len = nums.len();
+        for i in 0..len {
+            if nums[i] > 0 { break; }
+            if i > 0 && nums[i] == nums[i - 1] { continue; }
+            let mut set = HashSet::new();
+            for j in (i + 1)..len {
+                if j > i + 2 && nums[j] == nums[j - 1] && nums[j] == nums[j - 2] { continue; }
+                let c = 0 - (nums[i] + nums[j]);
+                if set.contains(&c) {
+                    result.push(vec![nums[i], nums[j], c]);
+                    set.remove(&c);
+                } else { set.insert(nums[j]); }
+            }
+        }
+        result
+    }
+}
+```
+
+```Rust
+// 双指针法
+use std::collections::HashSet;
+impl Solution {
+    pub fn three_sum(nums: Vec<i32>) -> Vec<Vec<i32>> {
+        let mut result: Vec<Vec<i32>> = Vec::new();
+        let mut nums = nums;
+        nums.sort();
+        let len = nums.len();
+        for i in 0..len {
+            if nums[i] > 0 { return result; }
+            if i > 0 && nums[i] == nums[i - 1] { continue; }
+            let (mut left, mut right) = (i + 1, len - 1);
+            while left < right {
+                if nums[i] + nums[left] + nums[right] > 0 {
+                    right -= 1;
+                    // 去重
+                    while left < right && nums[right] == nums[right + 1] { right -= 1; }
+                } else if nums[i] + nums[left] + nums[right] < 0 {
+                    left += 1;
+                    // 去重
+                    while left < right && nums[left] == nums[left - 1] { left += 1; }
+                } else {
+                    result.push(vec![nums[i], nums[left], nums[right]]);
+                    // 去重
+                    right -= 1;
+                    left += 1;
+                    while left < right && nums[right] == nums[right + 1] { right -= 1; }
+                    while left < right && nums[left] == nums[left - 1] { left += 1; }
+                }
+            }
+        }
+        result
+    }
+}
+```
+
 C:
 ```C
 //qsort辅助cmp函数

problems/0017.电话号码的字母组合.md

@@ -454,6 +454,49 @@ function letterCombinations(digits: string): string[] {
 };
 ```
 
+## Rust
+
+```Rust
+impl Solution {
+    fn backtracking(result: &mut Vec<String>, s: &mut String, map: &[&str; 10], digits: &String, index: usize)  {
+        let len = digits.len();
+        if len == index {
+            result.push(s.to_string());
+            return;
+        }
+        // 在保证不会越界的情况下使用unwrap()将Some()中的值提取出来
+        let digit= digits.chars().nth(index).unwrap().to_digit(10).unwrap() as usize;
+        let letters = map[digit];
+        for i in letters.chars() {
+            s.push(i);
+            Self::backtracking(result, s, &map, &digits, index+1);
+            s.pop();
+        }
+    }
+    pub fn letter_combinations(digits: String) -> Vec<String> {
+        if digits.len() == 0 {
+            return vec![];
+        }
+        const MAP: [&str; 10] = [
+        "",
+        "",
+        "abc",
+        "def",
+        "ghi",
+        "jkl",
+        "mno",
+        "pqrs",
+        "tuv",
+        "wxyz"
+        ];
+        let mut result: Vec<String> = Vec::new();
+        let mut s: String = String::new();
+        Self::backtracking(&mut result, &mut s, &MAP, &digits, 0);
+        result
+    }
+}
+```
+
 ## C
 
 ```c
@@ -557,6 +600,37 @@ func letterCombinations(_ digits: String) -> [String] {
 }
 ```
 
+## Scala:
+
+```scala
+object Solution {
+  import scala.collection.mutable
+  def letterCombinations(digits: String): List[String] = {
+    var result = mutable.ListBuffer[String]()
+    if(digits == "") return result.toList // 如果参数为空，返回空结果集的List形式
+    var path = mutable.ListBuffer[Char]()
+    // 数字和字符的映射关系
+    val map = Array[String]("", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz")
+
+    def backtracking(index: Int): Unit = {
+      if (index == digits.size) {
+        result.append(path.mkString)  // mkString语法：将数组类型直接转换为字符串
+        return
+      }
+      var digit = digits(index) - '0' // 这里使用toInt会报错！必须 -'0'
+      for (i <- 0 until map(digit).size) {
+        path.append(map(digit)(i))
+        backtracking(index + 1)
+        path = path.take(path.size - 1)
+      }
+    }
+    
+    backtracking(0)
+    result.toList
+  }
+}
+```
+
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0018.四数之和.md

@@ -135,6 +135,11 @@ class Solution {
        
         for (int i = 0; i < nums.length; i++) {
 
+            // nums[i] > target 直接返回, 剪枝操作
+            if (nums[i] > 0 && nums[i] > target) {
+                return result;
+            }
+
             if (i > 0 && nums[i - 1] == nums[i]) {
                 continue;
             }
@@ -148,7 +153,7 @@ class Solution {
                 int left = j + 1;
                 int right = nums.length - 1;
                 while (right > left) {
-                    int sum = nums[i] + nums[j] + nums[left] + nums[right];
+                    long sum = (long) nums[i] + nums[j] + nums[left] + nums[right];
                     if (sum > target) {
                         right--;
                     } else if (sum < target) {
@@ -517,6 +522,51 @@ public class Solution
     }
 }
 ```
+
+Rust:
+```Rust
+impl Solution {
+    pub fn four_sum(nums: Vec<i32>, target: i32) -> Vec<Vec<i32>> {
+        let mut result: Vec<Vec<i32>> = Vec::new();
+        let mut nums = nums;
+        nums.sort();
+        let len = nums.len();
+        for k in 0..len {
+            // 剪枝
+            if nums[k] > target && (nums[k] > 0 || target > 0) { break; }
+            // 去重
+            if k > 0 && nums[k] == nums[k - 1] { continue; }
+            for i in (k + 1)..len {
+                // 剪枝
+                if nums[k] + nums[i] > target && (nums[k] + nums[i] >= 0 || target >= 0) { break; }
+                // 去重
+                if i > k + 1 && nums[i] == nums[i - 1] { continue; }
+                let (mut left, mut right) = (i + 1, len - 1);
+                while left < right {
+                    if nums[k] + nums[i] > target - (nums[left] + nums[right]) {
+                        right -= 1;
+                        // 去重
+                        while left < right && nums[right] == nums[right + 1] { right -= 1; }
+                    } else if nums[k] + nums[i] < target - (nums[left] + nums[right]) {
+                        left += 1;
+                        // 去重
+                        while left < right && nums[left] == nums[left - 1] { left += 1; }
+                    } else {
+                        result.push(vec![nums[k], nums[i], nums[left], nums[right]]);
+                        // 去重
+                        while left < right && nums[right] == nums[right - 1] { right -= 1; }
+                        while left < right && nums[left] == nums[left + 1] { left += 1; }
+                        left += 1;
+                        right -= 1;
+                    }
+                }
+            }
+        }
+        result
+    }
+}
+```
+
 Scala:
 ```scala
 object Solution {

problems/0020.有效的括号.md

@@ -400,6 +400,37 @@ bool isValid(char * s){
     return !stackTop;
 }
 ```
+
+
+PHP:
+```php
+// https://www.php.net/manual/zh/class.splstack.php
+class Solution
+{
+    function isValid($s){
+        $stack = new SplStack();
+        for ($i = 0; $i < strlen($s); $i++) {
+            if ($s[$i] == "(") {
+                $stack->push(')');
+            } else if ($s[$i] == "{") {
+                $stack->push('}');
+            } else if ($s[$i] == "[") {
+                $stack->push(']');
+            // 2、遍历匹配过程中，发现栈内没有要匹配的字符 return false
+            // 3、遍历匹配过程中，栈已为空，没有匹配的字符了，说明右括号没有找到对应的左括号 return false
+            } else if ($stack->isEmpty() || $stack->top() != $s[$i]) {
+                return false;
+            } else {//$stack->top() == $s[$i]
+                $stack->pop();
+            }
+        }
+        // 1、遍历完，但是栈不为空,说明有相应的括号没有被匹配,return false
+        return $stack->isEmpty();
+    }
+}
+```
+
+
 Scala:
 ```scala
 object Solution {
@@ -422,5 +453,6 @@ object Solution {
   }
 }
 ```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0024.两两交换链表中的节点.md

@@ -256,20 +256,19 @@ TypeScript：
 
 ```typescript
 function swapPairs(head: ListNode | null): ListNode | null {
-  const dummyHead: ListNode = new ListNode(0, head);
-  let cur: ListNode = dummyHead;
-  while(cur.next !== null && cur.next.next !== null) {
-    const tem: ListNode = cur.next;
-    const tem1: ListNode = cur.next.next.next;
-    
-    cur.next = cur.next.next; // step 1
-    cur.next.next = tem; // step 2
-    cur.next.next.next = tem1; // step 3
-    
-    cur = cur.next.next;
-  }
-  return dummyHead.next;
-}
+    const dummyNode: ListNode = new ListNode(0, head);
+    let curNode: ListNode | null = dummyNode;
+    while (curNode && curNode.next && curNode.next.next) {
+        let firstNode: ListNode = curNode.next,
+            secNode: ListNode = curNode.next.next,
+            thirdNode: ListNode | null = curNode.next.next.next;
+        curNode.next = secNode;
+        secNode.next = firstNode;
+        firstNode.next = thirdNode;
+        curNode = firstNode;
+    }
+    return dummyNode.next;
+};
 ```
 
 Kotlin:

problems/0027.移除元素.md

@@ -219,6 +219,7 @@ func removeElement(nums []int, val int) int {
             res++
         }
     }
+    nums=nums[:res]
     return res
 }
 ```
@@ -339,7 +340,6 @@ int removeElement(int* nums, int numsSize, int val){
 }
 ```
 
-
 Kotlin:
 ```kotlin
 fun removeElement(nums: IntArray, `val`: Int): Int {
@@ -351,7 +351,6 @@ fun removeElement(nums: IntArray, `val`: Int): Int {
     }
 ```
 
-
 Scala:
 ```scala
 object Solution {
@@ -368,5 +367,20 @@ object Solution {
 }
 ```
 
+C#:
+```csharp
+public class Solution {
+    public int RemoveElement(int[] nums, int val) {
+        int slow = 0;
+        for (int fast = 0; fast < nums.Length; fast++) {
+            if (val != nums[fast]) {
+                nums[slow++] = nums[fast];
+            }
+        }
+        return slow;
+    }
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0028.实现strStr.md

@@ -1241,5 +1241,49 @@ function getNext(&$next, $s){
     }
 }
 ```
+
+Rust:
+
+> 前缀表统一不减一
+```Rust
+impl Solution {
+    pub fn get_next(next: &mut Vec<usize>, s: &Vec<char>) {
+        let len = s.len();
+        let mut j = 0;
+        for i in 1..len {
+            while j > 0 && s[i] != s[j] {
+                j = next[j - 1];
+            }
+            if s[i] == s[j] {
+                j += 1;
+            }
+            next[i] = j;
+        }
+    }
+
+    pub fn str_str(haystack: String, needle: String) -> i32 {
+        let (haystack_len, needle_len) = (haystack.len(), needle.len());
+        if haystack_len == 0 { return 0; }
+        if haystack_len < needle_len { return -1;}
+        let (haystack, needle) = (haystack.chars().collect::<Vec<char>>(), needle.chars().collect::<Vec<char>>());
+        let mut next: Vec<usize> = vec![0; haystack_len];
+        Self::get_next(&mut next, &needle);
+        let mut j = 0;
+        for i in 0..haystack_len {
+            while j > 0 && haystack[i] != needle[j] {
+                j = next[j - 1];
+            }
+            if haystack[i] == needle[j] {
+                j += 1;
+            }
+            if j == needle_len {
+                return (i - needle_len + 1) as i32;
+            }
+        }
+        return -1;
+    }
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0031.下一个排列.md

@@ -171,13 +171,13 @@ class Solution(object):
         i = n-2
         while i >= 0 and nums[i] >= nums[i+1]:
             i -= 1
-        
+
         if i > -1: // i==-1,不存在下一个更大的排列
             j = n-1
             while j >= 0 and nums[j] <= nums[i]:
                 j -= 1
             nums[i], nums[j] = nums[j], nums[i]
-
+    
         start, end = i+1, n-1
         while start < end:
             nums[start], nums[end] = nums[end], nums[start]
@@ -190,6 +190,26 @@ class Solution(object):
 ## Go
 
 ```go
+//卡尔的解法
+func nextPermutation(nums []int)  {
+    for i:=len(nums)-1;i>=0;i--{
+        for j:=len(nums)-1;j>i;j--{
+            if nums[j]>nums[i]{
+                //交换
+                nums[j],nums[i]=nums[i],nums[j]
+                reverse(nums,0+i+1,len(nums)-1)
+                return 
+            }
+        }
+    }
+    reverse(nums,0,len(nums)-1)
+}
+//对目标切片指定区间的反转方法
+func reverse(a []int,begin,end int){
+    for i,j:=begin,end;i<j;i,j=i+1,j-1{
+        a[i],a[j]=a[j],a[i]
+    }
+}
 ```
 
 ## JavaScript

problems/0034.在排序数组中查找元素的第一个和最后一个位置.md

@@ -482,6 +482,62 @@ var searchRange = function(nums, target) {
     return [-1, -1];
 };
 ```
+
+
+### TypeScript
+
+```typescript
+function searchRange(nums: number[], target: number): number[] {
+    const leftBoard: number = getLeftBorder(nums, target);
+    const rightBoard: number = getRightBorder(nums, target);
+    // target 在nums区间左侧或右侧
+    if (leftBoard === (nums.length - 1) || rightBoard === 0) return [-1, -1];
+    // target 不存在与nums范围内
+    if (rightBoard - leftBoard <= 1) return [-1, -1];
+    // target 存在于nums范围内
+    return [leftBoard + 1, rightBoard - 1];
+};
+// 查找第一个大于target的元素下标
+function getRightBorder(nums: number[], target: number): number {
+    let left: number = 0,
+        right: number = nums.length - 1;
+    // 0表示target在nums区间的左边
+    let rightBoard: number = 0;
+    while (left <= right) {
+        let mid = Math.floor((left + right) / 2);
+        if (nums[mid] <= target) {
+            // 右边界一定在mid右边（不含mid）
+            left = mid + 1;
+            rightBoard = left;
+        } else {
+            // 右边界在mid左边（含mid）
+            right = mid - 1;
+        }
+    }
+    return rightBoard;
+}
+// 查找第一个小于target的元素下标
+function getLeftBorder(nums: number[], target: number): number {
+    let left: number = 0,
+        right: number = nums.length - 1;
+    // length-1表示target在nums区间的右边
+    let leftBoard: number = nums.length - 1;
+    while (left <= right) {
+        let mid = Math.floor((left + right) / 2);
+        if (nums[mid] >= target) {
+            // 左边界一定在mid左边（不含mid）
+            right = mid - 1;
+            leftBoard = right;
+        } else {
+            // 左边界在mid右边（含mid）
+            left = mid + 1;
+        }
+    }
+    return leftBoard;
+}
+```
+
+
 ### Scala
 ```scala
 object Solution {
@@ -529,5 +585,49 @@ object Solution {
 }
 ```
 
+
+### Kotlin
+```kotlin
+class Solution {
+    fun searchRange(nums: IntArray, target: Int): IntArray {
+        var index = binarySearch(nums, target)
+        // 没找到，返回[-1, -1]
+        if (index == -1) return intArrayOf(-1, -1)
+        var left = index
+        var right = index
+        // 寻找左边界
+        while (left - 1 >=0 && nums[left - 1] == target){
+            left--
+        }
+        // 寻找右边界
+        while (right + 1 <nums.size && nums[right + 1] == target){
+            right++
+        }
+        return intArrayOf(left, right)
+    }
+    // 二分查找常规写法
+    fun binarySearch(nums: IntArray, target: Int): Int {
+        var left = 0;
+        var right = nums.size - 1
+        while (left <= right) {
+            var middle = left + (right - left)/2
+            if (nums[middle] > target) {
+                right = middle - 1
+            }
+            else {
+                if (nums[middle] < target) {
+                    left = middle + 1
+                }
+                else {
+                    return middle
+                }
+            }
+        }
+        // 没找到，返回-1
+        return -1
+    }
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0035.搜索插入位置.md

@@ -283,6 +283,28 @@ var searchInsert = function (nums, target) {
 };
 ```
 
+### TypeScript
+
+```typescript
+// 第一种二分法
+function searchInsert(nums: number[], target: number): number {
+    const length: number = nums.length;
+    let left: number = 0,
+        right: number = length - 1;
+    while (left <= right) {
+        const mid: number = Math.floor((left + right) / 2);
+        if (nums[mid] < target) {
+            left = mid + 1;
+        } else if (nums[mid] === target) {
+            return mid;
+        } else {
+            right = mid - 1;
+        }
+    }
+    return right + 1;
+};
+```
+
 ### Swift 
 
 ```swift

problems/0037.解数独.md

@@ -602,5 +602,100 @@ func solveSudoku(_ board: inout [[Character]]) {
 }
 ```
 
+### Scala
+
+详细写法：
+```scala
+object Solution {
+  
+  def solveSudoku(board: Array[Array[Char]]): Unit = {
+    backtracking(board)
+  }
+
+  def backtracking(board: Array[Array[Char]]): Boolean = {
+    for (i <- 0 until 9) {
+      for (j <- 0 until 9) {
+        if (board(i)(j) == '.') { // 必须是为 . 的数字才放数字
+          for (k <- '1' to '9') { // 这个位置放k是否合适
+            if (isVaild(i, j, k, board)) {
+              board(i)(j) = k
+              if (backtracking(board)) return true // 找到了立刻返回
+              board(i)(j) = '.' // 回溯
+            }
+          }
+          return false // 9个数都试完了，都不行就返回false
+        }
+      }
+    }
+    true // 遍历完所有的都没返回false，说明找到了
+  }
+
+  def isVaild(x: Int, y: Int, value: Char, board: Array[Array[Char]]): Boolean = {
+    // 行
+    for (i <- 0 until 9 ) {
+      if (board(i)(y) == value) {
+        return false
+      }
+    }
+    
+    // 列
+    for (j <- 0 until 9) {
+      if (board(x)(j) == value) {
+        return false
+      }
+    }
+
+    // 宫
+    var row = (x / 3) * 3
+    var col = (y / 3) * 3
+    for (i <- row until row + 3) {
+      for (j <- col until col + 3) {
+        if (board(i)(j) == value) {
+          return false
+        }
+      }
+    }
+
+    true
+  }
+}
+```
+
+遵循Scala至简原则写法：
+```scala
+object Solution {
+  
+  def solveSudoku(board: Array[Array[Char]]): Unit = {
+    backtracking(board)
+  }
+
+  def backtracking(board: Array[Array[Char]]): Boolean = {
+    // 双重for循环 + 循环守卫
+    for (i <- 0 until 9; j <- 0 until 9 if board(i)(j) == '.') {
+      // 必须是为 . 的数字才放数字，使用循环守卫判断该位置是否可以放置当前循环的数字
+      for (k <- '1' to '9' if isVaild(i, j, k, board)) { // 这个位置放k是否合适
+        board(i)(j) = k
+        if (backtracking(board)) return true // 找到了立刻返回
+        board(i)(j) = '.' // 回溯
+      }
+      return false // 9个数都试完了，都不行就返回false
+    }
+    true // 遍历完所有的都没返回false，说明找到了
+  }
+
+  def isVaild(x: Int, y: Int, value: Char, board: Array[Array[Char]]): Boolean = {
+    // 行，循环守卫进行判断
+    for (i <- 0 until 9 if board(i)(y) == value) return false
+    // 列，循环守卫进行判断
+    for (j <- 0 until 9 if board(x)(j) == value) return false
+    // 宫，循环守卫进行判断
+    var row = (x / 3) * 3
+    var col = (y / 3) * 3
+    for (i <- row until row + 3; j <- col until col + 3 if board(i)(j) == value) return false
+    true // 最终没有返回false，就说明该位置可以填写true
+  }
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0039.组合总和.md

@@ -291,7 +291,7 @@ class Solution:
         for i in range(start_index, len(candidates)):
             sum_ += candidates[i]
             self.path.append(candidates[i])
-            self.backtracking(candidates, target, sum_, i)  # 因为无限制重复选取，所以不是i-1
+            self.backtracking(candidates, target, sum_, i)  # 因为无限制重复选取，所以不是i+1
             sum_ -= candidates[i]   # 回溯
             self.path.pop()        # 回溯
 ```
@@ -502,5 +502,35 @@ func combinationSum(_ candidates: [Int], _ target: Int) -> [[Int]] {
 }
 ```
 
+## Scala
+
+```scala
+object Solution {
+  import scala.collection.mutable
+  def combinationSum(candidates: Array[Int], target: Int): List[List[Int]] = {
+    var result = mutable.ListBuffer[List[Int]]()
+    var path = mutable.ListBuffer[Int]()
+
+    def backtracking(sum: Int, index: Int): Unit = {
+      if (sum == target) {
+        result.append(path.toList) // 如果正好等于target，就添加到结果集
+        return
+      }
+      // 应该是从当前索引开始的，而不是从0
+      // 剪枝优化：添加循环守卫，当sum + c(i) <= target的时候才循环，才可以进入下一次递归
+      for (i <- index until candidates.size if sum + candidates(i) <= target) {
+        path.append(candidates(i))
+        backtracking(sum + candidates(i), i)
+        path = path.take(path.size - 1)
+      }
+    }
+
+    backtracking(0, 0)
+    result.toList
+  }
+}
+```
+
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0040.组合总和II.md

@@ -693,5 +693,37 @@ func combinationSum2(_ candidates: [Int], _ target: Int) -> [[Int]] {
 }
 ```
 
+
+## Scala
+
+```scala
+object Solution {
+  import scala.collection.mutable    
+  def combinationSum2(candidates: Array[Int], target: Int): List[List[Int]] = {
+    var res = mutable.ListBuffer[List[Int]]()
+    var path = mutable.ListBuffer[Int]()
+    var candidate = candidates.sorted
+
+    def backtracking(sum: Int, startIndex: Int): Unit = {
+      if (sum == target) {
+        res.append(path.toList)
+        return
+      }
+
+      for (i <- startIndex until candidate.size if sum + candidate(i) <= target) {
+        if (!(i > startIndex && candidate(i) == candidate(i - 1))) {
+          path.append(candidate(i))
+          backtracking(sum + candidate(i), i + 1)
+          path = path.take(path.size - 1)
+        }
+      }
+    }
+
+    backtracking(0, 0)
+    res.toList
+  }
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0042.接雨水.md

@@ -640,8 +640,44 @@ func min(a,b int)int{
 }
 ```
 
+单调栈解法
+```go
+func trap(height []int) int {
+   if len(height) <= 2 {
+      return 0
+   }
+   st := make([]int, 1, len(height)) // 切片模拟单调栈，st存储的是高度数组下标
+   var res int
+   for i := 1; i < len(height); i++ {
+      if height[i] < height[st[len(st)-1]] {
+         st = append(st, i)
+      } else if height[i] == height[st[len(st)-1]] {
+         st = st[:len(st)-1] // 比较的新元素和栈顶的元素相等，去掉栈中的，入栈新元素下标
+         st = append(st, i)
+      } else {
+         for len(st) != 0 && height[i] > height[st[len(st)-1]] {
+            top := st[len(st)-1]
+            st = st[:len(st)-1]
+            if len(st) != 0 {
+               tmp := (min(height[i], height[st[len(st)-1]]) - height[top]) * (i - st[len(st)-1] - 1)
+               res += tmp
+            }
+         }
+         st = append(st, i)
+      }
+   }
+   return res
+}
 
 
+func min(x, y int) int {
+   if x >= y {
+      return y
+   }
+   return x
+}
+```
+
 ### JavaScript:
 
 ```javascript

problems/0045.跳跃游戏II.md

@@ -279,7 +279,31 @@ function jump(nums: number[]): number {
 };
 ```
 
+### Scala
 
+```scala
+object Solution {
+  def jump(nums: Array[Int]): Int = {
+    if (nums.length == 0) return 0
+    var result = 0 // 记录走的最大步数
+    var curDistance = 0 // 当前覆盖最远距离下标
+    var nextDistance = 0 // 下一步覆盖最远距离下标
+    for (i <- nums.indices) {
+      nextDistance = math.max(nums(i) + i, nextDistance) // 更新下一步覆盖最远距离下标
+      if (i == curDistance) {
+        if (curDistance != nums.length - 1) {
+          result += 1
+          curDistance = nextDistance
+          if (nextDistance >= nums.length - 1) return result
+        } else {
+          return result
+        }
+      }
+    }
+    result
+  }
+}
+```
 
 
 

problems/0046.全排列.md

@@ -456,6 +456,36 @@ func permute(_ nums: [Int]) -> [[Int]] {
 }
 ```
 
+### Scala
+
+```scala
+object Solution {
+  import scala.collection.mutable
+  def permute(nums: Array[Int]): List[List[Int]] = {
+    var result = mutable.ListBuffer[List[Int]]()
+    var path = mutable.ListBuffer[Int]()
+
+    def backtracking(used: Array[Boolean]): Unit = {
+      if (path.size == nums.size) {
+        // 如果path的长度和nums相等，那么可以添加到结果集
+        result.append(path.toList)
+        return
+      }
+      // 添加循环守卫，只有当当前数字没有用过的情况下才进入回溯
+      for (i <- nums.indices if used(i) == false) {
+        used(i) = true
+        path.append(nums(i))
+        backtracking(used) // 回溯
+        path.remove(path.size - 1)
+        used(i) = false
+      }
+    }
+
+    backtracking(new Array[Boolean](nums.size)) // 调用方法
+    result.toList // 最终返回结果集的List形式
+  }
+}
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0047.全排列II.md

@@ -422,5 +422,43 @@ int** permuteUnique(int* nums, int numsSize, int* returnSize, int** returnColumn
 }
 ```
 
+### Scala
+
+```scala
+object Solution {
+  import scala.collection.mutable
+  def permuteUnique(nums: Array[Int]): List[List[Int]] = {
+    var result = mutable.ListBuffer[List[Int]]()
+    var path = mutable.ListBuffer[Int]()
+    var num = nums.sorted // 首先对数据进行排序
+
+    def backtracking(used: Array[Boolean]): Unit = {
+      if (path.size == num.size) {
+        // 如果path的size等于num了，那么可以添加到结果集
+        result.append(path.toList)
+        return
+      }
+      // 循环守卫，当前元素没被使用过就进入循环体
+      for (i <- num.indices if used(i) == false) {
+        // 当前索引为0，不存在和前一个数字相等可以进入回溯
+        // 当前索引值和上一个索引不相等，可以回溯
+        // 前一个索引对应的值没有被选，可以回溯
+        // 因为Scala没有continue，只能将逻辑反过来写 
+        if (i == 0 || (i > 0 && num(i) != num(i - 1)) || used(i-1) == false) {
+          used(i) = true
+          path.append(num(i))
+          backtracking(used)
+          path.remove(path.size - 1)
+          used(i) = false
+        }
+      }
+    }
+
+    backtracking(new Array[Boolean](nums.length))
+    result.toList
+  }
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0051.N皇后.md

@@ -455,7 +455,7 @@ var solveNQueens = function(n) {
 };
 ```
 
-## TypeScript
+### TypeScript
 
 ```typescript
 function solveNQueens(n: number): string[][] {
@@ -683,5 +683,77 @@ char *** solveNQueens(int n, int* returnSize, int** returnColumnSizes){
 }
 ```
 
+### Scala
+
+```scala
+object Solution {
+  import scala.collection.mutable
+  def solveNQueens(n: Int): List[List[String]] = {
+    var result = mutable.ListBuffer[List[String]]()
+
+    def judge(x: Int, y: Int, maze: Array[Array[Boolean]]): Boolean = {
+      // 正上方
+      var xx = x
+      while (xx >= 0) {
+        if (maze(xx)(y)) return false
+        xx -= 1
+      }
+      // 左边
+      var yy = y
+      while (yy >= 0) {
+        if (maze(x)(yy)) return false
+        yy -= 1
+      }
+      // 左上方
+      xx = x
+      yy = y
+      while (xx >= 0 && yy >= 0) {
+        if (maze(xx)(yy)) return false
+        xx -= 1
+        yy -= 1
+      }
+      xx = x
+      yy = y
+      // 右上方
+      while (xx >= 0 && yy < n) {
+        if (maze(xx)(yy)) return false
+        xx -= 1
+        yy += 1
+      }
+      true
+    }
+
+    def backtracking(row: Int, maze: Array[Array[Boolean]]): Unit = {
+      if (row == n) {
+        // 将结果转换为题目所需要的形式
+        var path = mutable.ListBuffer[String]()
+        for (x <- maze) {
+          var tmp = mutable.ListBuffer[String]()
+          for (y <- x) {
+            if (y == true) tmp.append("Q")
+            else tmp.append(".")
+          }
+          path.append(tmp.mkString)
+        }
+        result.append(path.toList)
+        return
+      }
+
+      for (j <- 0 until n) {
+        // 判断这个位置是否可以放置皇后
+        if (judge(row, j, maze)) {
+          maze(row)(j) = true
+          backtracking(row + 1, maze)
+          maze(row)(j) = false
+        }
+      }
+    }
+
+    backtracking(0, Array.ofDim[Boolean](n, n))
+    result.toList
+  }
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0053.最大子序和.md

@@ -333,8 +333,41 @@ function maxSubArray(nums: number[]): number {
 };
 ```
 
+### Scala
 
+**贪心**
 
+```scala
+object Solution {
+  def maxSubArray(nums: Array[Int]): Int = {
+    var result = Int.MinValue
+    var count = 0
+    for (i <- nums.indices) {
+      count += nums(i) // count累加
+      if (count > result) result = count // 记录最大值
+      if (count <= 0) count = 0 // 一旦count为负，则count归0
+    }
+    result
+  }
+}
+```
+
+**动态规划**
+
+```scala
+object Solution {
+  def maxSubArray(nums: Array[Int]): Int = {
+    var dp = new Array[Int](nums.length)
+    var result = nums(0)
+    dp(0) = nums(0)
+    for (i <- 1 until nums.length) {
+      dp(i) = math.max(nums(i), dp(i - 1) + nums(i))
+      result = math.max(result, dp(i))  // 更新最大值
+    }
+    result
+  }
+}
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0053.最大子序和（动态规划）.md

@@ -186,6 +186,24 @@ const maxSubArray = nums => {
 };
 ```
 
+
+Scala:
+
+```scala
+object Solution {
+  def maxSubArray(nums: Array[Int]): Int = {
+    var dp = new Array[Int](nums.length)
+    var result = nums(0)
+    dp(0) = nums(0)
+    for (i <- 1 until nums.length) {
+      dp(i) = math.max(nums(i), dp(i - 1) + nums(i)) 
+      result = math.max(result, dp(i))  // 更新最大值
+    }
+    result
+  }
+}
+```
+
 TypeScript：
 
 ```typescript
@@ -205,6 +223,5 @@ function maxSubArray(nums: number[]): number {
 ```
 
 
-
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0055.跳跃游戏.md

@@ -193,7 +193,22 @@ function canJump(nums: number[]): boolean {
 };
 ```
 
-
+### Scala
+```scala
+object Solution {
+  def canJump(nums: Array[Int]): Boolean = {
+    var cover = 0
+    if (nums.length == 1) return true // 如果只有一个元素，那么必定到达
+    var i = 0
+    while (i <= cover) { // i表示下标，当前只能够走cover步
+      cover = math.max(i + nums(i), cover)
+      if (cover >= nums.length - 1) return true // 说明可以覆盖到终点，直接返回
+      i += 1
+    }
+    false // 如果上面没有返回就是跳不到
+  }
+}
+```
 
 
 -----------------------

problems/0056.合并区间.md

@@ -133,24 +133,38 @@ public:
 
 ### Java 
 ```java
+
+/**
+时间复杂度 ： O(NlogN) 排序需要O(NlogN)
+空间复杂度 ： O(logN)  java 的内置排序是快速排序 需要 O(logN)空间
+
+*/
 class Solution {
     public int[][] merge(int[][] intervals) {
         List<int[]> res = new LinkedList<>();
-        Arrays.sort(intervals, (o1, o2) -> Integer.compare(o1[0], o2[0]));
-
+        //按照左边界排序
+        Arrays.sort(intervals, (x, y) -> Integer.compare(x[0], y[0]));
+        //initial start 是最小左边界
         int start = intervals[0][0];
+        int rightmostRightBound = intervals[0][1];
         for (int i = 1; i < intervals.length; i++) {
-            if (intervals[i][0] > intervals[i - 1][1]) {
-                res.add(new int[]{start, intervals[i - 1][1]});
+            //如果左边界大于最大右边界
+            if (intervals[i][0] > rightmostRightBound) {
+                //加入区间 并且更新start
+                res.add(new int[]{start, rightmostRightBound});
                 start = intervals[i][0];
+                rightmostRightBound = intervals[i][1];
             } else {
-                intervals[i][1] = Math.max(intervals[i][1], intervals[i - 1][1]);
+                //更新最大右边界
+                rightmostRightBound = Math.max(rightmostRightBound, intervals[i][1]);
             }
         }
-        res.add(new int[]{start, intervals[intervals.length - 1][1]});
+        res.add(new int[]{start, rightmostRightBound});
         return res.toArray(new int[res.size()][]);
     }
 }
+
+}
 ```
 ```java
 // 版本2

problems/0062.不同路径.md

@@ -374,6 +374,30 @@ function uniquePaths(m: number, n: number): number {
 };
 ```
 
+### Rust
+
+```Rust
+impl Solution {
+    pub fn unique_paths(m: i32, n: i32) -> i32 {
+            let m = m as usize;
+            let n = n as usize;
+            let mut dp = vec![vec![0; n]; m];
+            for i in 0..m {
+                dp[i][0] = 1;
+            }
+            for j in 0..n {
+                dp[0][j] = 1;
+            }
+            for i in 1..m {
+                for j in 1..n {
+                    dp[i][j] = dp[i-1][j] + dp[i][j-1];
+                }
+            }
+        dp[m-1][n-1]
+    }
+}
+```
+
 ### C
 
 ```c

problems/0063.不同路径II.md

@@ -384,6 +384,42 @@ function uniquePathsWithObstacles(obstacleGrid: number[][]): number {
 };
 ```
 
+### Rust
+
+```Rust
+impl Solution {
+    pub fn unique_paths_with_obstacles(obstacle_grid: Vec<Vec<i32>>) -> i32 {
+        let m: usize = obstacle_grid.len();
+        let n: usize = obstacle_grid[0].len();
+        if obstacle_grid[0][0] == 1 || obstacle_grid[m-1][n-1] == 1 {
+            return 0;
+        }
+        let mut dp = vec![vec![0; n]; m];
+        for i in 0..m {
+            if obstacle_grid[i][0] == 1 {
+                break;
+            }
+            else { dp[i][0] = 1; }
+        }
+        for j in 0..n {
+            if obstacle_grid[0][j] == 1 {
+                break;
+            }
+            else { dp[0][j] = 1; }
+        }
+        for i in 1..m {
+            for j in 1..n {
+                if obstacle_grid[i][j] == 1 {
+                    continue;
+                }
+                dp[i][j] = dp[i-1][j] + dp[i][j-1];
+            }
+        }
+        dp[m-1][n-1]
+    }
+}
+```
+
 ### C
 
 ```c

problems/0077.组合.md

@@ -535,6 +535,56 @@ func backtrack(n,k,start int,track []int){
 }
 ```
 
+### Rust
+
+```Rust
+impl Solution {
+    fn backtracking(result: &mut Vec<Vec<i32>>, path: &mut Vec<i32>, n: i32, k: i32, startIndex: i32) {
+        let len= path.len() as i32;
+        if len == k{
+            result.push(path.to_vec());
+            return;
+        }
+        for i in startIndex..= n {
+            path.push(i);
+            Self::backtracking(result, path, n, k, i+1);
+            path.pop();
+        }
+    }
+    pub fn combine(n: i32, k: i32) -> Vec<Vec<i32>> {
+        let mut result: Vec<Vec<i32>> = Vec::new();
+        let mut path: Vec<i32> = Vec::new();
+        Self::backtracking(&mut result, &mut path, n, k, 1);
+        result
+    }
+}
+```
+
+剪枝
+```Rust
+impl Solution {
+    fn backtracking(result: &mut Vec<Vec<i32>>, path: &mut Vec<i32>, n: i32, k: i32, startIndex: i32) {
+        let len= path.len() as i32;
+        if len == k{
+            result.push(path.to_vec());
+            return;
+        }
+	// 此处剪枝
+        for i in startIndex..= n - (k - len) + 1 {
+            path.push(i);
+            Self::backtracking(result, path, n, k, i+1);
+            path.pop();
+        }
+    }
+    pub fn combine(n: i32, k: i32) -> Vec<Vec<i32>> {
+        let mut result: Vec<Vec<i32>> = Vec::new();
+        let mut path: Vec<i32> = Vec::new();
+        Self::backtracking(&mut result, &mut path, n, k, 1);
+        result
+    }
+}
+```
+
 ### C
 ```c
 int* path;
@@ -673,5 +723,63 @@ func combine(_ n: Int, _ k: Int) -> [[Int]] {
 }
 ```
 
+### Scala
+
+暴力:
+```scala
+object Solution {
+  import scala.collection.mutable // 导包
+  def combine(n: Int, k: Int): List[List[Int]] = {
+    var result = mutable.ListBuffer[List[Int]]() // 存放结果集
+    var path = mutable.ListBuffer[Int]() //存放符合条件的结果
+    
+    def backtracking(n: Int, k: Int, startIndex: Int): Unit = {
+      if (path.size == k) {
+        // 如果path的size == k就达到题目要求，添加到结果集，并返回
+        result.append(path.toList)
+        return
+      }
+      for (i <- startIndex to n) { // 遍历从startIndex到n
+        path.append(i) // 先把数字添加进去
+        backtracking(n, k, i + 1) // 进行下一步回溯
+        path = path.take(path.size - 1) // 回溯完再删除掉刚刚添加的数字
+      }
+    }
+    
+    backtracking(n, k, 1) // 执行回溯
+    result.toList // 最终返回result的List形式，return关键字可以省略
+  }
+}
+```
+
+剪枝:
+
+```scala
+object Solution {
+  import scala.collection.mutable // 导包
+  def combine(n: Int, k: Int): List[List[Int]] = {
+    var result = mutable.ListBuffer[List[Int]]() // 存放结果集
+    var path = mutable.ListBuffer[Int]() //存放符合条件的结果
+
+    def backtracking(n: Int, k: Int, startIndex: Int): Unit = {
+      if (path.size == k) {
+        // 如果path的size == k就达到题目要求，添加到结果集，并返回
+        result.append(path.toList)
+        return
+      }
+      // 剪枝优化
+      for (i <- startIndex to (n - (k - path.size) + 1)) { 
+        path.append(i) // 先把数字添加进去
+        backtracking(n, k, i + 1) // 进行下一步回溯
+        path = path.take(path.size - 1) // 回溯完再删除掉刚刚添加的数字
+      }
+    }
+
+    backtracking(n, k, 1) // 执行回溯
+    result.toList // 最终返回result的List形式，return关键字可以省略
+  }
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0077.组合优化.md

@@ -261,6 +261,32 @@ function combine(n: number, k: number): number[][] {
 };
 ```
 
+Rust:
+
+```Rust
+impl Solution {
+    fn backtracking(result: &mut Vec<Vec<i32>>, path: &mut Vec<i32>, n: i32, k: i32, startIndex: i32) {
+        let len= path.len() as i32;
+        if len == k{
+            result.push(path.to_vec());
+            return;
+        }
+	// 此处剪枝
+        for i in startIndex..= n - (k - len) + 1 {
+            path.push(i);
+            Self::backtracking(result, path, n, k, i+1);
+            path.pop();
+        }
+    }
+    pub fn combine(n: i32, k: i32) -> Vec<Vec<i32>> {
+        let mut result: Vec<Vec<i32>> = Vec::new();
+        let mut path: Vec<i32> = Vec::new();
+        Self::backtracking(&mut result, &mut path, n, k, 1);
+        result
+    }
+}
+```
+
 C:
 
 ```c
@@ -346,5 +372,34 @@ func combine(_ n: Int, _ k: Int) -> [[Int]] {
 }
 ```
 
+Scala:
+
+```scala
+object Solution {
+  import scala.collection.mutable // 导包
+  def combine(n: Int, k: Int): List[List[Int]] = {
+    var result = mutable.ListBuffer[List[Int]]() // 存放结果集
+    var path = mutable.ListBuffer[Int]() //存放符合条件的结果
+
+    def backtracking(n: Int, k: Int, startIndex: Int): Unit = {
+      if (path.size == k) {
+        // 如果path的size == k就达到题目要求，添加到结果集，并返回
+        result.append(path.toList)
+        return
+      }
+      // 剪枝优化
+      for (i <- startIndex to (n - (k - path.size) + 1)) { 
+        path.append(i) // 先把数字添加进去
+        backtracking(n, k, i + 1) // 进行下一步回溯
+        path = path.take(path.size - 1) // 回溯完再删除掉刚刚添加的数字
+      }
+    }
+
+    backtracking(n, k, 1) // 执行回溯
+    result.toList // 最终返回result的List形式，return关键字可以省略
+  }
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0078.子集.md

@@ -373,6 +373,60 @@ func subsets(_ nums: [Int]) -> [[Int]] {
 }
 ```
 
+## Scala
+
+思路一: 使用本题解思路
+
+```scala
+object Solution {
+  import scala.collection.mutable
+  def subsets(nums: Array[Int]): List[List[Int]] = {
+    var result = mutable.ListBuffer[List[Int]]()
+    var path = mutable.ListBuffer[Int]()
+
+    def backtracking(startIndex: Int): Unit = {
+      result.append(path.toList) // 存放结果
+      if (startIndex >= nums.size) {
+        return
+      }
+      for (i <- startIndex until nums.size) {
+        path.append(nums(i)) // 添加元素
+        backtracking(i + 1)
+        path.remove(path.size - 1) // 删除
+      }
+    }
+
+    backtracking(0)
+    result.toList
+  }
+}
+```
+
+思路二: 将原问题转换为二叉树，针对每一个元素都有**选或不选**两种选择，直到遍历到最后，所有的叶子节点即为本题的答案：
+
+```scala
+object Solution {
+  import scala.collection.mutable
+  def subsets(nums: Array[Int]): List[List[Int]] = {
+    var result = mutable.ListBuffer[List[Int]]()
+
+    def backtracking(path: mutable.ListBuffer[Int], startIndex: Int): Unit = {
+      if (startIndex == nums.length) {
+        result.append(path.toList)
+        return
+      }
+      path.append(nums(startIndex))
+      backtracking(path, startIndex + 1)  // 选择元素
+      path.remove(path.size - 1)
+      backtracking(path, startIndex + 1)  // 不选择元素
+    }
+
+    backtracking(mutable.ListBuffer[Int](), 0)
+    result.toList
+  }
+}
+```
+
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0090.子集II.md

@@ -434,6 +434,63 @@ func subsetsWithDup(_ nums: [Int]) -> [[Int]] {
 }
 ```
 
+### Scala
+
+不使用userd数组:
+
+```scala
+object Solution {
+  import scala.collection.mutable
+  def subsetsWithDup(nums: Array[Int]): List[List[Int]] = {
+    var result = mutable.ListBuffer[List[Int]]()
+    var path = mutable.ListBuffer[Int]()
+    var num = nums.sorted // 排序
+
+    def backtracking(startIndex: Int): Unit = {
+      result.append(path.toList)
+      if (startIndex >= num.size){
+        return
+      }
+      for (i <- startIndex until num.size) {
+        // 同一树层重复的元素不进入回溯
+        if (!(i > startIndex && num(i) == num(i - 1))) {
+          path.append(num(i))
+          backtracking(i + 1)
+          path.remove(path.size - 1)
+        }
+      }
+    }
+
+    backtracking(0)
+    result.toList
+  }
+}
+```
+
+使用Set去重:
+```scala
+object Solution {
+  import scala.collection.mutable
+  def subsetsWithDup(nums: Array[Int]): List[List[Int]] = {
+    var result = mutable.Set[List[Int]]()
+    var num = nums.sorted
+    def backtracking(path: mutable.ListBuffer[Int], startIndex: Int): Unit = {
+      if (startIndex == num.length) {
+        result.add(path.toList)
+        return
+      }
+      path.append(num(startIndex))
+      backtracking(path, startIndex + 1)  // 选择
+      path.remove(path.size - 1)
+      backtracking(path, startIndex + 1)  // 不选择
+    }
+
+    backtracking(mutable.ListBuffer[Int](), 0)
+
+    result.toList
+  }
+}
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0093.复原IP地址.md

@@ -659,6 +659,48 @@ func restoreIpAddresses(_ s: String) -> [String] {
 }
 ```
 
+## Scala
+
+```scala
+object Solution {
+  import scala.collection.mutable
+  def restoreIpAddresses(s: String): List[String] = {
+    var result = mutable.ListBuffer[String]()
+    if (s.size < 4 || s.length > 12) return result.toList
+    var path = mutable.ListBuffer[String]()
+
+    // 判断IP中的一个字段是否为正确的
+    def isIP(sub: String): Boolean = {
+      if (sub.size > 1 && sub(0) == '0') return false
+      if (sub.toInt > 255) return false
+      true
+    }
+
+    def backtracking(startIndex: Int): Unit = {
+      if (startIndex >= s.size) {
+        if (path.size == 4) {
+          result.append(path.mkString(".")) // mkString方法可以把集合里的数据以指定字符串拼接
+          return
+        }
+        return
+      }
+      // subString
+      for (i <- startIndex until startIndex + 3 if i < s.size) {
+        var subString = s.substring(startIndex, i + 1)
+        if (isIP(subString)) { // 如果合法则进行下一轮
+          path.append(subString)
+          backtracking(i + 1)
+          path = path.take(path.size - 1)
+        }
+      }
+    }
+
+    backtracking(0)
+    result.toList
+  }
+}
+```
+
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0098.验证二叉搜索树.md

@@ -589,7 +589,50 @@ function isValidBST(root: TreeNode | null): boolean {
 };
 ```
 
+## Scala
 
+辅助数组解决:
+```scala
+object Solution {
+  import scala.collection.mutable
+  def isValidBST(root: TreeNode): Boolean = {
+    var arr = new mutable.ArrayBuffer[Int]()
+    // 递归中序遍历二叉树，将节点添加到arr
+    def traversal(node: TreeNode): Unit = {
+      if (node == null) return
+      traversal(node.left)
+      arr.append(node.value)
+      traversal(node.right)
+    }
+    traversal(root)
+    // 这个数组如果是升序就代表是二叉搜索树
+    for (i <- 1 until arr.size) {
+      if (arr(i) <= arr(i - 1)) return false
+    }
+    true  
+  }
+}
+```
+
+递归中解决:
+```scala
+object Solution {
+  def isValidBST(root: TreeNode): Boolean = {
+    var flag = true
+    var preValue:Long = Long.MinValue // 这里要使用Long类型
+
+    def traversal(node: TreeNode): Unit = {
+      if (node == null || flag == false) return
+      traversal(node.left)
+      if (node.value > preValue) preValue = node.value
+      else flag = false
+      traversal(node.right)
+    }
+    traversal(root)
+    flag  
+  }
+}
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0100.相同的树.md

@@ -240,6 +240,46 @@ Go：
 
 JavaScript：
 
+TypeScript:
+
+> 递归法-先序遍历
+
+```typescript
+function isSameTree(p: TreeNode | null, q: TreeNode | null): boolean {
+    if (p === null && q === null) return true;
+    if (p === null || q === null) return false;
+    if (p.val !== q.val) return false;
+    return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
+};
+```
+
+> 迭代法-层序遍历
+
+```typescript
+function isSameTree(p: TreeNode | null, q: TreeNode | null): boolean {
+    const queue1: (TreeNode | null)[] = [],
+        queue2: (TreeNode | null)[] = [];
+    queue1.push(p);
+    queue2.push(q);
+    while (queue1.length > 0 && queue2.length > 0) {
+        const node1 = queue1.shift(),
+            node2 = queue2.shift();
+        if (node1 === null && node2 === null) continue;
+        if (
+            (node1 === null || node2 === null) ||
+            node1!.val !== node2!.val
+        ) return false;
+        queue1.push(node1!.left);
+        queue1.push(node1!.right);
+        queue2.push(node2!.left);
+        queue2.push(node2!.right);
+    }
+    return true;
+};
+```
+
+
+
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0101.对称二叉树.md

@@ -437,6 +437,31 @@ class Solution:
         return True
 ```
 
+层次遍历
+```python
+class Solution:
+    def isSymmetric(self, root: Optional[TreeNode]) -> bool:
+        if not root:
+            return True
+
+        que = [root]
+        while que:
+            this_level_length = len(que)
+            for i in range(this_level_length // 2):
+                # 要么其中一个是None但另外一个不是
+                if (not que[i] and que[this_level_length - 1 - i]) or (que[i] and not que[this_level_length - 1 - i]):
+                    return False
+                # 要么两个都不是None
+                if que[i] and que[i].val != que[this_level_length - 1 - i].val:
+                    return False
+            for i in range(this_level_length):
+                if not que[i]: continue
+                que.append(que[i].left)
+                que.append(que[i].right)
+            que = que[this_level_length:]
+        return True
+```
+
 ## Go
 
 ```go

problems/0102.二叉树的层序遍历.md

@@ -2629,21 +2629,21 @@ JavaScript：
 var minDepth = function(root) {
     if (root === null) return 0;
     let queue = [root];
-    let deepth = 0;
+    let depth = 0;
     while (queue.length) {
         let n = queue.length;
-        deepth++;
+        depth++;
         for (let i=0; i<n; i++) {
             let node = queue.shift();
-            // 如果左右节点都是null，则该节点深度最小
+            // 如果左右节点都是null(在遇见的第一个leaf节点上)，则该节点深度最小
             if (node.left === null && node.right === null) {
-                return deepth;
+                return depth;
             }
             node.left && queue.push(node.left);;
-            node.right && queue.push (node.right);
+            node.right && queue.push(node.right);
         }
     }
-    return deepth;
+    return depth;
 };
 ```
 

problems/0104.二叉树的最大深度.md

@@ -294,14 +294,13 @@ class solution {
     /**
      * 递归法
      */
-    public int maxdepth(treenode root) {
+    public int maxDepth(TreeNode root) {
         if (root == null) {
             return 0;
         }
-        int leftdepth = maxdepth(root.left);
-        int rightdepth = maxdepth(root.right);
-        return math.max(leftdepth, rightdepth) + 1;
-
+        int leftDepth = maxDepth(root.left);
+        int rightDepth = maxDepth(root.right);
+        return Math.max(leftDepth, rightDepth) + 1;
     }
 }
 ```
@@ -311,23 +310,23 @@ class solution {
     /**
      * 迭代法，使用层序遍历
      */
-    public int maxdepth(treenode root) {
+    public int maxDepth(TreeNode root) {
         if(root == null) {
             return 0;
         }
-        deque<treenode> deque = new linkedlist<>();
+        Deque<TreeNode> deque = new LinkedList<>();
         deque.offer(root);
         int depth = 0;
-        while (!deque.isempty()) {
+        while (!deque.isEmpty()) {
             int size = deque.size();
             depth++;
             for (int i = 0; i < size; i++) {
-                treenode poll = deque.poll();
-                if (poll.left != null) {
-                    deque.offer(poll.left);
+                TreeNode node = deque.poll();
+                if (node.left != null) {
+                    deque.offer(node.left);
                 }
-                if (poll.right != null) {
-                    deque.offer(poll.right);
+                if (node.right != null) {
+                    deque.offer(node.right);
                 }
             }
         }

problems/0106.从中序与后序遍历序列构造二叉树.md

@@ -584,35 +584,29 @@ tree2 的前序遍历是[1 2 3]， 后序遍历是[3 2 1]。
 
 ```java
 class Solution {
+    Map<Integer, Integer> map;  // 方便根据数值查找位置
     public TreeNode buildTree(int[] inorder, int[] postorder) {
-        return buildTree1(inorder, 0, inorder.length, postorder, 0, postorder.length);
+        map = new HashMap<>();
+        for (int i = 0; i < inorder.length; i++) { // 用map保存中序序列的数值对应位置
+            map.put(inorder[i], i);
+        }
+
+        return findNode(inorder,  0, inorder.length, postorder,0, postorder.length);  // 前闭后开
     }
-    public TreeNode buildTree1(int[] inorder, int inLeft, int inRight,
-                               int[] postorder, int postLeft, int postRight) {
-        // 没有元素了
-        if (inRight - inLeft < 1) {
+    
+    public TreeNode findNode(int[] inorder, int inBegin, int inEnd, int[] postorder, int postBegin, int postEnd) {
+        // 参数里的范围都是前闭后开
+        if (inBegin >= inEnd || postBegin >= postEnd) {  // 不满足左闭右开，说明没有元素，返回空树
             return null;
         }
-        // 只有一个元素了
-        if (inRight - inLeft == 1) {
-            return new TreeNode(inorder[inLeft]);
-        }
-        // 后序数组postorder里最后一个即为根结点
-        int rootVal = postorder[postRight - 1];
-        TreeNode root = new TreeNode(rootVal);
-        int rootIndex = 0;
-        // 根据根结点的值找到该值在中序数组inorder里的位置
-        for (int i = inLeft; i < inRight; i++) {
-            if (inorder[i] == rootVal) {
-                rootIndex = i;
-                break;
-            }
-        }
-        // 根据rootIndex划分左右子树
-        root.left = buildTree1(inorder, inLeft, rootIndex,
-                postorder, postLeft, postLeft + (rootIndex - inLeft));
-        root.right = buildTree1(inorder, rootIndex + 1, inRight,
-                postorder, postLeft + (rootIndex - inLeft), postRight - 1);
+        int rootIndex = map.get(postorder[postEnd - 1]);  // 找到后序遍历的最后一个元素在中序遍历中的位置
+        TreeNode root = new TreeNode(inorder[rootIndex]);  // 构造结点
+        int lenOfLeft = rootIndex - inBegin;  // 保存中序左子树个数，用来确定后序数列的个数
+        root.left = findNode(inorder, inBegin, rootIndex,
+                            postorder, postBegin, postBegin + lenOfLeft);
+        root.right = findNode(inorder, rootIndex + 1, inEnd,
+                            postorder, postBegin + lenOfLeft, postEnd - 1);
+
         return root;
     }
 }
@@ -622,31 +616,29 @@ class Solution {
 
 ```java
 class Solution {
+    Map<Integer, Integer> map;
     public TreeNode buildTree(int[] preorder, int[] inorder) {
-        return helper(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);
-    }
-
-    public TreeNode helper(int[] preorder, int preLeft, int preRight,
-                           int[] inorder, int inLeft, int inRight) {
-        // 递归终止条件
-        if (inLeft > inRight || preLeft > preRight) return null;
-
-        // val 为前序遍历第一个的值，也即是根节点的值
-        // idx 为根据根节点的值来找中序遍历的下标
-        int idx = inLeft, val = preorder[preLeft];
-        TreeNode root = new TreeNode(val);
-        for (int i = inLeft; i <= inRight; i++) {
-            if (inorder[i] == val) {
-                idx = i;
-                break;
-            }
+        map = new HashMap<>();
+        for (int i = 0; i < inorder.length; i++) { // 用map保存中序序列的数值对应位置
+            map.put(inorder[i], i);
         }
 
-        // 根据 idx 来递归找左右子树
-        root.left = helper(preorder, preLeft + 1, preLeft + (idx - inLeft),
-                         inorder, inLeft, idx - 1);
-        root.right = helper(preorder, preLeft + (idx - inLeft) + 1, preRight,
-                         inorder, idx + 1, inRight);
+        return findNode(preorder, 0, preorder.length, inorder,  0, inorder.length);  // 前闭后开
+    }
+
+    public TreeNode findNode(int[] preorder, int preBegin, int preEnd, int[] inorder, int inBegin, int inEnd) {
+        // 参数里的范围都是前闭后开
+        if (preBegin >= preEnd || inBegin >= inEnd) {  // 不满足左闭右开，说明没有元素，返回空树
+            return null;
+        }
+        int rootIndex = map.get(preorder[preBegin]);  // 找到前序遍历的第一个元素在中序遍历中的位置
+        TreeNode root = new TreeNode(inorder[rootIndex]);  // 构造结点
+        int lenOfLeft = rootIndex - inBegin;  // 保存中序左子树个数，用来确定前序数列的个数
+        root.left = findNode(preorder, preBegin + 1, preBegin + lenOfLeft + 1,  
+                            inorder, inBegin, rootIndex);
+        root.right = findNode(preorder, preBegin + lenOfLeft + 1, preEnd,
+                            inorder, rootIndex + 1, inEnd);
+
         return root;
     }
 }

problems/0108.将有序数组转换为二叉搜索树.md

@@ -448,5 +448,27 @@ struct TreeNode* sortedArrayToBST(int* nums, int numsSize) {
 }
 ```
 
+## Scala
+
+递归:
+
+```scala
+object Solution {
+  def sortedArrayToBST(nums: Array[Int]): TreeNode = {
+    def buildTree(left: Int, right: Int): TreeNode = {
+      if (left > right) return null // 当left大于right的时候，返回空
+      // 最中间的节点是当前节点
+      var mid = left + (right - left) / 2
+      var curNode = new TreeNode(nums(mid))
+      curNode.left = buildTree(left, mid - 1)
+      curNode.right = buildTree(mid + 1, right)
+      curNode
+    }
+    buildTree(0, nums.size - 1)
+  }
+}
+```
+
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0110.平衡二叉树.md

@@ -531,40 +531,26 @@ class Solution:
 迭代法：
 ```python
 class Solution:
-    def isBalanced(self, root: TreeNode) -> bool:
-        st = []
+    def isBalanced(self, root: Optional[TreeNode]) -> bool:
         if not root:
             return True
-        st.append(root)
-        while st:
-            node = st.pop() #中
-            if abs(self.getDepth(node.left) - self.getDepth(node.right)) > 1:
-                return False
-            if node.right:
-                st.append(node.right) #右（空节点不入栈）
-            if node.left:
-                st.append(node.left) #左（空节点不入栈）
-        return True
-    
-    def getDepth(self, cur):
-        st = []
-        if cur:
-            st.append(cur)
-        depth = 0
-        result = 0
-        while st:
-            node = st.pop()
+
+        height_map = {}
+        stack = [root]
+        while stack:
+            node = stack.pop()
             if node:
-                st.append(node) #中
-                st.append(None)
-                depth += 1
-                if node.right: st.append(node.right) #右
-                if node.left: st.append(node.left) #左
+                stack.append(node)
+                stack.append(None)
+                if node.left: stack.append(node.left)
+                if node.right: stack.append(node.right)
             else:
-                node = st.pop()
-                depth -= 1
-            result = max(result, depth)
-        return result
+                real_node = stack.pop()
+                left, right = height_map.get(real_node.left, 0), height_map.get(real_node.right, 0)
+                if abs(left - right) > 1:
+                    return False
+                height_map[real_node] = 1 + max(left, right)
+        return True
 ```
 
 

problems/0116.填充每个节点的下一个右侧节点指针.md

@@ -287,6 +287,79 @@ const connect = root => {
 };
 ```
 
+## TypeScript
+
+（注：命名空间‘Node’与typescript中内置类型冲突，这里改成了‘NodePro’）
+
+> 递归法：
+
+```typescript
+class NodePro {
+    val: number
+    left: NodePro | null
+    right: NodePro | null
+    next: NodePro | null
+    constructor(val?: number, left?: NodePro, right?: NodePro, next?: NodePro) {
+        this.val = (val === undefined ? 0 : val)
+        this.left = (left === undefined ? null : left)
+        this.right = (right === undefined ? null : right)
+        this.next = (next === undefined ? null : next)
+    }
+}
+
+function connect(root: NodePro | null): NodePro | null {
+    if (root === null) return null;
+    root.next = null;
+    recur(root);
+    return root;
+};
+function recur(node: NodePro): void {
+    if (node.left === null || node.right === null) return;
+    node.left.next = node.right;
+    node.right.next = node.next && node.next.left;
+    recur(node.left);
+    recur(node.right);
+}
+```
+
+> 迭代法：
+
+```typescript
+class NodePro {
+    val: number
+    left: NodePro | null
+    right: NodePro | null
+    next: NodePro | null
+    constructor(val?: number, left?: NodePro, right?: NodePro, next?: NodePro) {
+        this.val = (val === undefined ? 0 : val)
+        this.left = (left === undefined ? null : left)
+        this.right = (right === undefined ? null : right)
+        this.next = (next === undefined ? null : next)
+    }
+}
+
+function connect(root: NodePro | null): NodePro | null {
+    if (root === null) return null;
+    const queue: NodePro[] = [];
+    queue.push(root);
+    while (queue.length > 0) {
+        for (let i = 0, length = queue.length; i < length; i++) {
+            const curNode: NodePro = queue.shift()!;
+            if (i === length - 1) {
+                curNode.next = null;
+            } else {
+                curNode.next = queue[0];
+            }
+            if (curNode.left !== null) queue.push(curNode.left);
+            if (curNode.right !== null) queue.push(curNode.right);
+        }
+    }
+    return root;
+};
+```
+
+
+
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0122.买卖股票的最佳时机II.md

@@ -133,8 +133,9 @@ public:
 
 ## 其他语言版本
 
-Java: 
+### Java: 
 
+贪心:
 ```java
 // 贪心思路
 class Solution {
@@ -148,6 +149,7 @@ class Solution {
 }
 ```
 
+动态规划:
 ```java
 class Solution { // 动态规划
     public int maxProfit(int[] prices) {
@@ -169,8 +171,8 @@ class Solution { // 动态规划
 }
 ```
 
-Python: 
-
+### Python: 
+贪心:
 ```python
 class Solution:
     def maxProfit(self, prices: List[int]) -> int:
@@ -180,7 +182,7 @@ class Solution:
         return result
 ```
 
-python动态规划
+动态规划:
 ```python
 class Solution:
     def maxProfit(self, prices: List[int]) -> int:
@@ -194,7 +196,7 @@ class Solution:
         return dp[-1][1]
 ```
 
-Go:
+### Go:
 
 ```golang
 //贪心算法
@@ -231,7 +233,7 @@ func maxProfit(prices []int) int {
 }
 ```
 
-Javascript:
+### Javascript:
 
 贪心
 ```Javascript
@@ -268,7 +270,7 @@ const maxProfit = (prices) => {
 };
 ```
 
-TypeScript：
+### TypeScript：
 
 ```typescript
 function maxProfit(prices: number[]): number {
@@ -280,7 +282,7 @@ function maxProfit(prices: number[]): number {
 };
 ```
 
-C:
+### C:
 贪心：
 ```c
 int maxProfit(int* prices, int pricesSize){
@@ -318,5 +320,22 @@ int maxProfit(int* prices, int pricesSize){
 }
 ```
 
+### Scala
+
+贪心:
+```scala
+object Solution {
+  def maxProfit(prices: Array[Int]): Int = {
+    var result = 0
+    for (i <- 1 until prices.length) {
+      if (prices(i) > prices(i - 1)) {
+        result += prices(i) - prices(i - 1)
+      }
+    }
+    result
+  }
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0129.求根到叶子节点数字之和.md

@@ -3,6 +3,9 @@
   <img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20210924105952.png" width="1000"/>
 </a>
 <p align="center"><strong><a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！</strong></p>
+
+
+
 # 129. 求根节点到叶节点数字之和
 
 [力扣题目链接](https://leetcode.cn/problems/sum-root-to-leaf-numbers/)
@@ -245,6 +248,29 @@ class Solution:
 ```
 Go：
 
+```go
+func sumNumbers(root *TreeNode) int {
+	sum = 0
+	travel(root, root.Val)
+	return sum
+}
+
+func travel(root *TreeNode, tmpSum int) {
+	if root.Left == nil && root.Right == nil {
+		sum += tmpSum
+	} else {
+		if root.Left != nil {
+			travel(root.Left, tmpSum*10+root.Left.Val)
+		}
+		if root.Right != nil {
+			travel(root.Right, tmpSum*10+root.Right.Val)
+		}
+	}
+}
+```
+
+
+
 JavaScript：
 ```javascript
 var sumNumbers = function(root) {
@@ -289,7 +315,40 @@ var sumNumbers = function(root) {
 };
 ```
 
+TypeScript：
+
+```typescript
+function sumNumbers(root: TreeNode | null): number {
+    if (root === null) return 0;
+    let resTotal: number = 0;
+    const route: number[] = [];
+    route.push(root.val);
+    recur(root, route);
+    return resTotal;
+    function recur(node: TreeNode, route: number[]): void {
+        if (node.left === null && node.right === null) {
+            resTotal += listToSum(route);
+            return;
+        }
+        if (node.left !== null) {
+            route.push(node.left.val);
+            recur(node.left, route);
+            route.pop();
+        };
+        if (node.right !== null) {
+            route.push(node.right.val);
+            recur(node.right, route);
+            route.pop();
+        };
+    }
+    function listToSum(nums: number[]): number {
+        return Number(nums.join(''));
+    }
+};
+```
+
 C:
+
 ```c
 //sum记录总和
 int sum;

problems/0131.分割回文串.md

@@ -676,5 +676,50 @@ impl Solution {
     }
 }
 ```
+
+
+## Scala
+
+```scala
+object Solution {
+
+  import scala.collection.mutable
+
+  def partition(s: String): List[List[String]] = {
+    var result = mutable.ListBuffer[List[String]]()
+    var path = mutable.ListBuffer[String]()
+
+    // 判断字符串是否回文
+    def isPalindrome(start: Int, end: Int): Boolean = {
+      var (left, right) = (start, end)
+      while (left < right) {
+        if (s(left) != s(right)) return false
+        left += 1
+        right -= 1
+      }
+      true
+    }
+
+    // 回溯算法
+    def backtracking(startIndex: Int): Unit = {
+      if (startIndex >= s.size) {
+        result.append(path.toList)
+        return
+      }
+      // 添加循环守卫，如果当前分割是回文子串则进入回溯
+      for (i <- startIndex until s.size if isPalindrome(startIndex, i)) {
+        path.append(s.substring(startIndex, i + 1))
+        backtracking(i + 1)
+        path = path.take(path.size - 1)
+      }
+    }
+
+    backtracking(0)
+    result.toList
+  }
+}
+```
+
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0132.分割回文串II.md

@@ -206,6 +206,55 @@ public:
 ## Java
 
 ```java
+class Solution {
+
+    public int minCut(String s) {
+        if(null == s || "".equals(s)){
+            return 0;
+        }
+        int len = s.length();
+        // 1.
+        // 记录子串[i..j]是否是回文串
+        boolean[][] isPalindromic = new boolean[len][len];
+        // 从下到上，从左到右
+        for(int i = len - 1; i >= 0; i--){
+            for(int j = i; j < len; j++){
+                if(s.charAt(i) == s.charAt(j)){
+                    if(j - i <= 1){
+                        isPalindromic[i][j] = true;
+                    } else{
+                        isPalindromic[i][j] = isPalindromic[i + 1][j - 1];
+                    }
+                } else{
+                    isPalindromic[i][j] = false;
+                }
+            }
+        }
+
+        // 2.
+        // dp[i] 表示[0..i]的最小分割次数
+        int[] dp = new int[len];
+        for(int i = 0; i < len; i++){
+            //初始考虑最坏的情况。 1个字符分割0次， len个字符分割 len - 1次
+            dp[i] = i;
+        }
+
+        for(int i = 1; i < len; i++){
+            if(isPalindromic[0][i]){
+                // s[0..i]是回文了，那 dp[i] = 0, 一次也不用分割
+                dp[i] = 0;
+                continue;
+            }
+            for(int j = 0; j < i; j++){
+                // 按文中的思路，不清楚就拿 "ababa" 为例，先写出 isPalindromic 数组，再进行求解
+                if(isPalindromic[j + 1][i]){
+                    dp[i] = Math.min(dp[i], dp[j] + 1);
+                }
+            }
+        }
+        return dp[len - 1];
+    }
+}
 ```
 
 ## Python
@@ -240,6 +289,7 @@ class Solution:
 ## Go
 
 ```go
+
 ```
 
 ## JavaScript

problems/0141.环形链表.md

@@ -7,6 +7,8 @@
 
 # 141. 环形链表
 
+[力扣题目链接](https://leetcode.cn/problems/linked-list-cycle/submissions/)
+
 给定一个链表，判断链表中是否有环。
 
 如果链表中有某个节点，可以通过连续跟踪 next 指针再次到达，则链表中存在环。 为了表示给定链表中的环，我们使用整数 pos 来表示链表尾连接到链表中的位置（索引从 0 开始）。 如果 pos 是 -1，则在该链表中没有环。注意：pos 不作为参数进行传递，仅仅是为了标识链表的实际情况。
@@ -103,7 +105,7 @@ class Solution:
         return False
 ```
 
-## Go
+### Go
 
 ```go
 func hasCycle(head *ListNode) bool {
@@ -139,6 +141,23 @@ var hasCycle = function(head) {
 };
 ```
 
+### TypeScript
+
+```typescript
+function hasCycle(head: ListNode | null): boolean {
+    let slowNode: ListNode | null = head,
+        fastNode: ListNode | null = head;
+    while (fastNode !== null && fastNode.next !== null) {
+        slowNode = slowNode!.next;
+        fastNode = fastNode.next.next;
+        if (slowNode === fastNode) return true;
+    }
+    return false;
+};
+```
+
+
+
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0142.环形链表II.md

@@ -303,13 +303,13 @@ function detectCycle(head: ListNode | null): ListNode | null {
     let slowNode: ListNode | null = head,
         fastNode: ListNode | null = head;
     while (fastNode !== null && fastNode.next !== null) {
-        slowNode = (slowNode as ListNode).next;
+        slowNode = slowNode!.next;
         fastNode = fastNode.next.next;
         if (slowNode === fastNode) {
             slowNode = head;
             while (slowNode !== fastNode) {
-                slowNode = (slowNode as ListNode).next;
-                fastNode = (fastNode as ListNode).next;
+                slowNode = slowNode!.next;
+                fastNode = fastNode!.next;
             }
             return slowNode;
         }

problems/0143.重排链表.md

@@ -6,6 +6,8 @@
 
 # 143.重排链表
 
+[力扣题目链接](https://leetcode.cn/problems/reorder-list/submissions/)
+
 ![](https://code-thinking-1253855093.file.myqcloud.com/pics/20210726160122.png)
 
 ## 思路
@@ -465,7 +467,81 @@ var reorderList = function(head, s = [], tmp) {
 }
 ```
 
+### TypeScript
+
+> 辅助数组法：
+
+```typescript
+function reorderList(head: ListNode | null): void {
+    if (head === null) return;
+    const helperArr: ListNode[] = [];
+    let curNode: ListNode | null = head;
+    while (curNode !== null) {
+        helperArr.push(curNode);
+        curNode = curNode.next;
+    }
+    let node: ListNode = head;
+    let left: number = 1,
+        right: number = helperArr.length - 1;
+    let count: number = 0;
+    while (left <= right) {
+        if (count % 2 === 0) {
+            node.next = helperArr[right--];
+        } else {
+            node.next = helperArr[left++];
+        }
+        count++;
+        node = node.next;
+    }
+    node.next = null;
+};
+```
+
+> 分割链表法：
+
+```typescript
+function reorderList(head: ListNode | null): void {
+    if (head === null || head.next === null) return;
+    let fastNode: ListNode = head,
+        slowNode: ListNode = head;
+    while (fastNode.next !== null && fastNode.next.next !== null) {
+        slowNode = slowNode.next!;
+        fastNode = fastNode.next.next;
+    }
+    let head1: ListNode | null = head;
+    // 反转后半部分链表
+    let head2: ListNode | null = reverseList(slowNode.next);
+    // 分割链表
+    slowNode.next = null;
+    /**
+    	直接在head1链表上进行插入
+        head1 链表长度一定大于或等于head2,
+        因此在下面的循环中，只要head2不为null, head1 一定不为null
+     */
+    while (head2 !== null) {
+        const tempNode1: ListNode | null = head1!.next,
+            tempNode2: ListNode | null = head2.next;
+        head1!.next = head2;
+        head2.next = tempNode1;
+        head1 = tempNode1;
+        head2 = tempNode2;
+    }
+};
+function reverseList(head: ListNode | null): ListNode | null {
+    let curNode: ListNode | null = head,
+        preNode: ListNode | null = null;
+    while (curNode !== null) {
+        const tempNode: ListNode | null = curNode.next;
+        curNode.next = preNode;
+        preNode = curNode;
+        curNode = tempNode;
+    }
+    return preNode;
+}
+```
+
 ### C
+
 方法三：反转链表
 ```c
 //翻转链表

problems/0150.逆波兰表达式求值.md

@@ -325,6 +325,33 @@ func evalRPN(_ tokens: [String]) -> Int {
     return stack.last!
 }
 ```
+
+
+PHP：
+```php
+class Solution {
+    function evalRPN($tokens) {
+        $st = new SplStack();
+        for($i = 0;$i<count($tokens);$i++){
+            // 是数字直接入栈
+            if(is_numeric($tokens[$i])){ 
+                $st->push($tokens[$i]);
+            }else{ 
+                // 是符号进行运算
+                $num1 = $st->pop();
+                $num2 = $st->pop();
+                if ($tokens[$i] == "+") $st->push($num2 + $num1);
+                if ($tokens[$i] == "-") $st->push($num2 - $num1);
+                if ($tokens[$i] == "*") $st->push($num2 * $num1);
+                // 注意处理小数部分
+                if ($tokens[$i] == "/") $st->push(intval($num2 / $num1));
+            }
+        }
+        return $st->pop();
+    }
+}
+```
+
 Scala:
 ```scala
 object Solution {
@@ -351,6 +378,7 @@ object Solution {
     // 最后返回栈顶，不需要加return关键字
     stack.pop() 
   }
+
 }
 ```
 -----------------------

problems/0151.翻转字符串里的单词.md

@@ -79,7 +79,7 @@ void removeExtraSpaces(string& s) {
 
 逻辑很简单，从前向后遍历，遇到空格了就erase。
 
-如果不仔细琢磨一下erase的时间复杂读，还以为以上的代码是O(n)的时间复杂度呢。
+如果不仔细琢磨一下erase的时间复杂度，还以为以上的代码是O(n)的时间复杂度呢。
 
 想一下真正的时间复杂度是多少，一个erase本来就是O(n)的操作，erase实现原理题目：[数组：就移除个元素很难么？](https://programmercarl.com/0027.移除元素.html)，最优的算法来移除元素也要O(n)。
 
@@ -864,7 +864,58 @@ function reverseString(&$s, $start, $end) {
     return ;
 }
 ```
+Rust:
 
+```Rust
+// 根据C++版本二思路进行实现
+// 函数名根据Rust编译器建议由驼峰命名法改为蛇形命名法
+impl Solution {
+    pub fn reverse(s: &mut Vec<char>, mut begin: usize, mut end: usize){
+        while begin < end {
+            let temp = s[begin];
+            s[begin] = s[end];
+            s[end] = temp;
+            begin += 1;
+            end -= 1;
+        }
+}
+pub fn remove_extra_spaces(s: &mut Vec<char>) {
+        let mut slow: usize = 0;
+        let len = s.len();
+        // 注意这里不能用for循环，不然在里面那个while循环中对i的递增会失效
+        let mut i: usize = 0;
+        while i < len {
+            if !s[i].is_ascii_whitespace() {
+                if slow != 0 {
+                    s[slow] = ' ';
+                    slow += 1;
+                }
+                while i < len && !s[i].is_ascii_whitespace() {
+                    s[slow] = s[i];
+                    slow += 1;
+                    i += 1;
+                }
+            }
+            i += 1;
+        }
+        s.resize(slow, ' ');
+    }
+    pub fn reverse_words(s: String) -> String {
+        let mut s = s.chars().collect::<Vec<char>>();
+        Self::remove_extra_spaces(&mut s);
+        let len = s.len();
+        Self::reverse(&mut s, 0, len - 1);
+        let mut start = 0;
+        for i in 0..=len {
+            if i == len || s[i].is_ascii_whitespace() {
+                Self::reverse(&mut s, start, i - 1);
+                start = i + 1;
+            }
+        }
+        s.iter().collect::<String>()
+    }
+}
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0189.旋转数组.md

@@ -7,6 +7,8 @@
 
 # 189. 旋转数组
 
+[力扣题目链接](https://leetcode.cn/problems/rotate-array/)
+
 给定一个数组，将数组中的元素向右移动 k 个位置，其中 k 是非负数。
 
 进阶：
@@ -160,6 +162,27 @@ var rotate = function (nums, k) {
 };
 ```
 
+## TypeScript
+
+```typescript
+function rotate(nums: number[], k: number): void {
+    const length: number = nums.length;
+    k %= length;
+    reverseByRange(nums, 0, length - 1);
+    reverseByRange(nums, 0, k - 1);
+    reverseByRange(nums, k, length - 1);
+};
+function reverseByRange(nums: number[], left: number, right: number): void {
+    while (left < right) {
+        const temp = nums[left];
+        nums[left] = nums[right];
+        nums[right] = temp;
+        left++;
+        right--;
+    }
+}
+```
+
 
 
 -----------------------

problems/0202.快乐数.md

@@ -315,6 +315,36 @@ class Solution {
 }
 ```
 
+Rust:
+```Rust
+use std::collections::HashSet;
+impl Solution {
+    pub fn get_sum(mut n: i32) -> i32 {
+        let mut sum = 0;
+        while n > 0 {
+            sum += (n % 10) * (n % 10);
+            n /= 10;
+        }
+        sum
+    }
+
+    pub fn is_happy(n: i32) -> bool {
+        let mut n = n;
+        let mut set = HashSet::new();
+        loop {
+            let sum = Self::get_sum(n);
+            if sum == 1 {
+                return true;
+            }
+            if set.contains(&sum) {
+                return false;
+            } else { set.insert(sum); }
+            n = sum;
+        }
+    }
+}
+```
+
 C:
 ```C
 typedef struct HashNodeTag {

problems/0203.移除链表元素.md

@@ -397,18 +397,18 @@ function removeElements(head: ListNode | null, val: number): ListNode | null {
 
 ```typescript
 function removeElements(head: ListNode | null, val: number): ListNode | null {
-    let dummyHead = new ListNode(0, head);
-    let pre: ListNode = dummyHead, cur: ListNode | null = dummyHead.next;
-    // 删除非头部节点
+    // 添加虚拟节点
+    const data = new ListNode(0, head);
+    let pre = data, cur = data.next;
     while (cur) {
         if (cur.val === val) {
-            pre.next = cur.next;
+            pre.next = cur.next
         } else {
             pre = cur;
         }
         cur = cur.next;
     }
-    return head.next;
+    return data.next;
 };
 ```
 
@@ -487,17 +487,19 @@ RUST:
 // }
 impl Solution {
     pub fn remove_elements(head: Option<Box<ListNode>>, val: i32) -> Option<Box<ListNode>> {
-        let mut head = head;
-        let mut dummy_head = ListNode::new(0);
-        let mut cur = &mut dummy_head;
-        while let Some(mut node) = head {
-            head = std::mem::replace(&mut node.next, None);
-            if node.val != val {
-                cur.next = Some(node);
+        let mut dummyHead = Box::new(ListNode::new(0));
+        dummyHead.next = head;
+        let mut cur = dummyHead.as_mut();
+	// 使用take()替换std::men::replace(&mut node.next, None)达到相同的效果，并且更普遍易读
+        while let Some(nxt) = cur.next.take() {
+            if nxt.val == val {
+                cur.next = nxt.next;
+            } else {
+                cur.next = Some(nxt);
                 cur = cur.next.as_mut().unwrap();
             }
         }
-        dummy_head.next
+        dummyHead.next
     }
 }
 ```
@@ -532,5 +534,39 @@ object Solution {
   }
 }
 ```
+Kotlin:
+```kotlin
+/**
+ * Example:
+ * var li = ListNode(5)
+ * var v = li.`val`
+ * Definition for singly-linked list.
+ * class ListNode(var `val`: Int) {
+ *     var next: ListNode? = null
+ * }
+ */
+class Solution {
+    fun removeElements(head: ListNode?, `val`: Int): ListNode? {
+        // 使用虚拟节点，令该节点指向head
+        var dummyNode = ListNode(-1)
+        dummyNode.next = head
+        // 使用cur遍历链表各个节点
+        var cur = dummyNode
+        // 判断下个节点是否为空
+        while (cur.next != null) {
+            // 符合条件，移除节点
+            if (cur.next.`val` == `val`) {
+                cur.next = cur.next.next
+            }
+            // 不符合条件，遍历下一节点
+            else {
+                cur = cur.next
+            }
+        }
+        // 注意：返回的不是虚拟节点
+        return dummyNode.next
+    }
+}
+```
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0205.同构字符串.md

@@ -156,6 +156,28 @@ var isIsomorphic = function(s, t) {
 };
 ```
 
+## TypeScript
+
+```typescript
+function isIsomorphic(s: string, t: string): boolean {
+    const helperMap1: Map<string, string> = new Map();
+    const helperMap2: Map<string, string> = new Map();
+    for (let i = 0, length = s.length; i < length; i++) {
+        let temp1: string | undefined = helperMap1.get(s[i]);
+        let temp2: string | undefined = helperMap2.get(t[i]);
+        if (temp1 === undefined && temp2 === undefined) {
+            helperMap1.set(s[i], t[i]);
+            helperMap2.set(t[i], s[i]);
+        } else if (temp1 !== t[i] || temp2 !== s[i]) {
+            return false;
+        }
+    }
+    return true;
+};
+```
+
+
+
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0206.翻转链表.md

@@ -422,6 +422,41 @@ fun reverseList(head: ListNode?): ListNode? {
     return pre
 }
 ```
+```kotlin
+/**
+ * Example:
+ * var li = ListNode(5)
+ * var v = li.`val`
+ * Definition for singly-linked list.
+ * class ListNode(var `val`: Int) {
+ *     var next: ListNode? = null
+ * }
+ */
+class Solution {
+    fun reverseList(head: ListNode?): ListNode? {
+        // temp用来存储临时的节点
+        var temp: ListNode?
+        // cur用来遍历链表
+        var cur: ListNode? = head
+        // pre用来作为链表反转的工具
+        // pre是比pre前一位的节点
+        var pre: ListNode? = null
+        while (cur != null) {
+            // 临时存储原本cur的下一个节点
+            temp = cur.next
+            // 使cur下一节点地址为它之前的
+            cur.next = pre
+            // 之后随着cur的遍历移动pre
+            pre = cur;
+            // 移动cur遍历链表各个节点
+            cur = temp;
+        }
+        // 由于开头使用pre为null,所以cur等于链表本身长度+1，
+        // 此时pre在cur前一位，所以此时pre为头节点
+        return pre;
+    }
+}
+```
 
 Swift：
 ```swift

problems/0209.长度最小的子数组.md

@@ -417,6 +417,38 @@ class Solution {
     }
 }
 ```
+滑动窗口
+```kotlin
+class Solution {
+   fun minSubArrayLen(target: Int, nums: IntArray): Int {
+    // 左边界 和 右边界
+    var left: Int = 0
+    var right: Int = 0
+    // sum 用来记录和
+    var sum: Int = 0
+    // result记录一个固定值，便于判断是否存在的这样的数组
+    var result: Int = Int.MAX_VALUE
+    // subLenth记录长度
+    var subLength = Int.MAX_VALUE
+   
+
+    while (right < nums.size) {
+        // 从数组首元素开始逐次求和
+        sum += nums[right++]
+        // 判断
+        while (sum >= target) {
+            var temp = right - left 
+            // 每次和上一次比较求出最小数组长度
+            subLength = if (subLength > temp) temp else subLength
+            // sum减少，左边界右移
+            sum -= nums[left++]
+        }
+    }
+    // 如果subLength为初始值，则说明长度为0，否则返回subLength
+    return if(subLength == result) 0 else subLength
+    }
+}
+```
 Scala:
 
 滑动窗口:
@@ -465,6 +497,27 @@ object Solution {
   }
 }
 ```
-
+C#:
+```csharp
+public class Solution {
+    public int MinSubArrayLen(int s, int[] nums) {
+        int n = nums.Length;
+        int ans = int.MaxValue;
+        int start = 0, end = 0;
+        int sum = 0;
+        while (end < n)  {
+            sum += nums[end];
+            while (sum >= s) 
+            {
+                ans = Math.Min(ans, end - start + 1);
+                sum -= nums[start];
+                start++;
+            }
+            end++;
+        }
+        return ans == int.MaxValue ? 0 : ans;
+    }
+}
+```
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0216.组合总和III.md

@@ -411,6 +411,35 @@ function combinationSum3(k: number, n: number): number[][] {
 };
 ```
 
+## Rust
+
+```Rust
+impl Solution {
+    fn backtracking(result: &mut Vec<Vec<i32>>, path:&mut Vec<i32>, targetSum:i32, k: i32, mut sum: i32, startIndex: i32) {
+        let len = path.len() as i32;
+        if len == k {
+            if sum == targetSum {
+                result.push(path.to_vec());
+            }
+            return;
+        }
+        for i in startIndex..=9 {
+            sum += i;
+            path.push(i);
+            Self::backtracking(result, path, targetSum, k, sum, i+1);
+            sum -= i;
+            path.pop();
+        }
+    }
+    pub fn combination_sum3(k: i32, n: i32) -> Vec<Vec<i32>> {
+        let mut result: Vec<Vec<i32>> = Vec::new();
+        let mut path: Vec<i32> = Vec::new();
+        Self::backtracking(&mut result, &mut path, n, k, 0, 1);
+        result
+    }
+}
+```
+
 ## C
 
 ```c
@@ -502,5 +531,35 @@ func combinationSum3(_ count: Int, _ targetSum: Int) -> [[Int]] {
 }
 ```
 
+## Scala
+
+```scala
+object Solution {
+  import scala.collection.mutable
+  def combinationSum3(k: Int, n: Int): List[List[Int]] = {
+    var result = mutable.ListBuffer[List[Int]]()
+    var path = mutable.ListBuffer[Int]()
+
+    def backtracking(k: Int, n: Int, sum: Int, startIndex: Int): Unit = {
+      if (sum > n) return // 剪枝，如果sum>目标和，就返回
+      if (sum == n && path.size == k) {
+        result.append(path.toList)
+        return
+      }
+      // 剪枝
+      for (i <- startIndex to (9 - (k - path.size) + 1)) {
+        path.append(i)
+        backtracking(k, n, sum + i, i + 1) 
+        path = path.take(path.size - 1)
+      }
+    }
+
+    backtracking(k, n, 0, 1) // 调用递归方法
+    result.toList // 最终返回结果集的List形式
+  }
+}
+```
+
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0225.用队列实现栈.md

@@ -816,7 +816,6 @@ class MyStack {
 }
 ```
 Scala:
-
 使用两个队列模拟栈:
 ```scala
 import scala.collection.mutable
@@ -897,6 +896,86 @@ class MyStack() {
   def empty(): Boolean = {
     queue.isEmpty
   }
+  }
+```
+
+
+PHP
+> 双对列
+```php
+// SplQueue 类通过使用一个双向链表来提供队列的主要功能。(PHP 5 >= 5.3.0, PHP 7, PHP 8)
+// https://www.php.net/manual/zh/class.splqueue.php
+class MyStack {
+    public $queueMain; // 保存数据
+    public $queueTmp;  // 辅助作用
+
+    function __construct() {
+        $this->queueMain=new SplQueue();
+        $this->queueTmp=new SplQueue();
+    }
+    
+    // queueMain: 1,2,3 <= add
+    function push($x) {
+        $this->queueMain->enqueue($x);
+    }
+                                    
+    function pop() {
+       $qmSize = $this->queueMain->Count();
+       $qmSize --;
+       // queueMain: 3,2,1 => pop =>2,1 => add => 2,1 :queueTmp
+       while($qmSize --){
+           $this->queueTmp->enqueue($this->queueMain->dequeue());
+       }
+       // queueMain: 3
+       $val = $this->queueMain->dequeue();
+       // queueMain <= queueTmp 
+       $this->queueMain = $this->queueTmp;
+       // 清空queueTmp,下次使用 
+       $this->queueTmp = new SplQueue();
+       return $val;
+    }
+
+    function top() {
+        // 底层是双链表实现：从双链表的末尾查看节点
+        return $this->queueMain->top(); 
+    }
+
+    function empty() {
+        return $this->queueMain->isEmpty();
+    }
+}
+```
+> 单对列
+```php
+class MyStack {
+    public $queue;
+
+    function __construct() {
+        $this->queue=new SplQueue();
+    }
+    
+    function push($x) {
+        $this->queue->enqueue($x);
+    }
+                                 
+    function pop() {
+       $qmSize = $this->queue->Count();
+       $qmSize --;
+       //queue: 3,2,1 => pop =>2,1 => add => 2,1,3 :queue
+       while($qmSize --){
+           $this->queue->enqueue($this->queue->dequeue());
+       }
+       $val = $this->queue->dequeue();
+       return $val;
+    }
+
+    function top() {
+        return $this->queue->top(); 
+    }
+
+    function empty() {
+        return $this->queue->isEmpty();
+    }
 }
 ```
 -----------------------

problems/0226.翻转二叉树.md

@@ -470,25 +470,14 @@ func invertTree(root *TreeNode) *TreeNode {
 使用递归版本的前序遍历
 ```javascript
 var invertTree = function(root) {
-    //1. 首先使用递归版本的前序遍历实现二叉树翻转
-    //交换节点函数
-    const inverNode=function(left,right){
-        let temp=left;
-        left=right;
-        right=temp;
-        //需要重新给root赋值一下
-        root.left=left;
-        root.right=right;
+    // 终止条件
+    if (!root) {
+        return null;
     }
-    //确定递归函数的参数和返回值inverTree=function(root)
-    //确定终止条件
-    if(root===null){
-        return root;
-    }
-    //确定节点处理逻辑 交换
-    inverNode(root.left,root.right);
-    invertTree(root.left);
-    invertTree(root.right);
+    // 交换左右节点
+    const rightNode = root.right;
+    root.right = invertTree(root.left);
+    root.left = invertTree(rightNode);
     return root;
 };
 ```

problems/0232.用栈实现队列.md

@@ -495,6 +495,53 @@ void myQueueFree(MyQueue* obj) {
     obj->stackOutTop = 0;
 }
 ```
+
+
+PHP:
+```php
+// SplStack 类通过使用一个双向链表来提供栈的主要功能。[PHP 5 >= 5.3.0, PHP 7, PHP 8]
+// https://www.php.net/manual/zh/class.splstack.php
+class MyQueue {
+    // 双栈模拟队列：In栈存储数据；Out栈辅助处理
+    private $stackIn;
+    private $stackOut;
+    
+    function __construct() {
+        $this->stackIn = new SplStack();
+        $this->stackOut = new SplStack();
+    }
+
+    // In: 1 2 3  <= push  
+    function push($x) {
+        $this->stackIn->push($x);
+    }
+
+    function pop() {
+        $this->peek();
+        return $this->stackOut->pop();
+    }
+
+    function peek() {
+        if($this->stackOut->isEmpty()){
+            $this->shift();
+        }
+        return $this->stackOut->top();
+    }
+
+    function empty() {
+        return $this->stackOut->isEmpty() && $this->stackIn->isEmpty();
+    }
+
+    // 如果Out栈为空，把In栈数据压入Out栈
+    // In: 1 2 3 => pop    push => 1 2 3 :Out  
+    private function shift(){
+        while(!$this->stackIn->isEmpty()){
+            $this->stackOut->push($this->stackIn->pop());
+        }
+    }
+    }
+```
+
 Scala:
 ```scala
 class MyQueue() {
@@ -533,6 +580,7 @@ class MyQueue() {
   def empty(): Boolean = {
     stackIn.isEmpty && stackOut.isEmpty
   }
+
 }
 ```
 -----------------------

problems/0234.回文链表.md

@@ -273,7 +273,7 @@ class Solution:
         return pre
 ```
 
-## Go
+### Go
 
 ```go
 
@@ -319,6 +319,63 @@ var isPalindrome = function(head) {
 };
 ```
 
+### TypeScript
+
+> 数组模拟
+
+```typescript
+function isPalindrome(head: ListNode | null): boolean {
+    const helperArr: number[] = [];
+    let curNode: ListNode | null = head;
+    while (curNode !== null) {
+        helperArr.push(curNode.val);
+        curNode = curNode.next;
+    }
+    let left: number = 0,
+        right: number = helperArr.length - 1;
+    while (left < right) {
+        if (helperArr[left++] !== helperArr[right--]) return false;
+    }
+    return true;
+};
+```
+
+> 反转后半部分链表
+
+```typescript
+function isPalindrome(head: ListNode | null): boolean {
+    if (head === null || head.next === null) return true;
+    let fastNode: ListNode | null = head,
+        slowNode: ListNode = head,
+        preNode: ListNode = head;
+    while (fastNode !== null && fastNode.next !== null) {
+        preNode = slowNode;
+        slowNode = slowNode.next!;
+        fastNode = fastNode.next.next;
+    }
+    preNode.next = null;
+    let cur1: ListNode | null = head;
+    let cur2: ListNode | null = reverseList(slowNode);
+    while (cur1 !== null) {
+        if (cur1.val !== cur2!.val) return false;
+        cur1 = cur1.next;
+        cur2 = cur2!.next;
+    }
+    return true;
+};
+function reverseList(head: ListNode | null): ListNode | null {
+    let curNode: ListNode | null = head,
+        preNode: ListNode | null = null;
+    while (curNode !== null) {
+        let tempNode: ListNode | null = curNode.next;
+        curNode.next = preNode;
+        preNode = curNode;
+        curNode = tempNode;
+    }
+    return preNode;
+}
+```
+
 
 
 -----------------------

problems/0235.二叉搜索树的最近公共祖先.md

@@ -381,7 +381,36 @@ function lowestCommonAncestor(root: TreeNode | null, p: TreeNode | null, q: Tree
 };
 ```
 
+## Scala
 
+递归:
+
+```scala
+object Solution {
+  def lowestCommonAncestor(root: TreeNode, p: TreeNode, q: TreeNode): TreeNode = {
+    // scala中每个关键字都有其返回值，于是可以不写return
+    if (root.value > p.value && root.value > q.value) lowestCommonAncestor(root.left, p, q)
+    else if (root.value < p.value && root.value < q.value) lowestCommonAncestor(root.right, p, q)
+    else root
+  }
+}
+```
+
+迭代:
+
+```scala
+object Solution {
+  def lowestCommonAncestor(root: TreeNode, p: TreeNode, q: TreeNode): TreeNode = {
+    var curNode = root  // 因为root是不可变量，所以要赋值给curNode一个可变量
+    while(curNode != null){
+      if(curNode.value > p.value && curNode.value > q.value) curNode = curNode.left
+      else if(curNode.value < p.value && curNode.value < q.value) curNode = curNode.right
+      else return curNode
+    }
+    null
+  }
+}
+```
 
 
 -----------------------

problems/0236.二叉树的最近公共祖先.md

@@ -343,7 +343,25 @@ function lowestCommonAncestor(root: TreeNode | null, p: TreeNode | null, q: Tree
 };
 ```
 
+## Scala
 
+```scala
+object Solution {
+  def lowestCommonAncestor(root: TreeNode, p: TreeNode, q: TreeNode): TreeNode = {
+    // 递归结束条件
+    if (root == null || root == p || root == q) {
+      return root
+    }
+
+    var left = lowestCommonAncestor(root.left, p, q)
+    var right = lowestCommonAncestor(root.right, p, q)
+
+    if (left != null && right != null) return root
+    if (left == null) return right
+    left
+  }
+}
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0239.滑动窗口最大值.md

@@ -654,8 +654,7 @@ object Solution {
     // 最终返回res，return关键字可以省略
     res
   }
-
-}
+ }
 
 class MyQueue {
   var queue = ArrayBuffer[Int]()
@@ -678,5 +677,84 @@ class MyQueue {
   def peek(): Int = queue.head
 }
 ```
+
+
+PHP:
+```php
+class Solution {
+    /**
+     * @param Integer[] $nums
+     * @param Integer $k
+     * @return Integer[]
+     */
+    function maxSlidingWindow($nums, $k) {
+        $myQueue = new MyQueue();
+        // 先将前k的元素放进队列
+        for ($i = 0; $i < $k; $i++) { 
+            $myQueue->push($nums[$i]);
+        }
+
+        $result = [];
+        $result[] = $myQueue->max(); // result 记录前k的元素的最大值
+
+        for ($i = $k; $i < count($nums); $i++) {
+            $myQueue->pop($nums[$i - $k]); // 滑动窗口移除最前面元素
+            $myQueue->push($nums[$i]); // 滑动窗口前加入最后面的元素
+            $result[]= $myQueue->max(); // 记录对应的最大值
+        }
+        return $result;
+    }
+
+}
+
+// 单调对列构建
+class MyQueue{
+    private $queue;
+
+    public function __construct(){
+        $this->queue = new SplQueue(); //底层是双向链表实现。
+    }
+
+    public function pop($v){
+        // 判断当前对列是否为空
+        // 比较当前要弹出的数值是否等于队列出口元素的数值，如果相等则弹出。
+        // bottom 从链表前端查看元素, dequeue 从双向链表的开头移动一个节点
+        if(!$this->queue->isEmpty() && $v == $this->queue->bottom()){
+            $this->queue->dequeue(); //弹出队列
+        }
+    }
+
+    public function push($v){
+        // 判断当前对列是否为空
+        // 如果push的数值大于入口元素的数值，那么就将队列后端的数值弹出，直到push的数值小于等于队列入口元素的数值为止。
+        // 这样就保持了队列里的数值是单调从大到小的了。
+        while (!$this->queue->isEmpty() && $v > $this->queue->top()) {
+            $this->queue->pop(); // pop从链表末尾弹出一个元素，
+        }
+        $this->queue->enqueue($v);
+    }
+
+    // 查询当前队列里的最大值 直接返回队首
+    public function max(){
+        // bottom 从链表前端查看元素, top从链表末尾查看元素
+        return $this->queue->bottom(); 
+    }
+
+    // 辅助理解: 打印队列元素
+    public function println(){
+        // "迭代器移动到链表头部": 可理解为从头遍历链表元素做准备。
+        // 【PHP中没有指针概念，所以就没说指针。从数据结构上理解，就是把指针指向链表头部】
+        $this->queue->rewind();
+
+        echo "Println: ";
+        while($this->queue->valid()){
+            echo $this->queue->current()," -> ";
+            $this->queue->next();
+        }
+        echo "\n";
+    } 
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0283.移动零.md

@@ -133,6 +133,27 @@ var moveZeroes = function(nums) {
 };
 ```
 
+TypeScript：
+
+```typescript
+function moveZeroes(nums: number[]): void {
+    const length: number = nums.length;
+    let slowIndex: number = 0,
+        fastIndex: number = 0;
+    while (fastIndex < length) {
+        if (nums[fastIndex] !== 0) {
+            nums[slowIndex++] = nums[fastIndex];
+        };
+        fastIndex++;
+    }
+    while (slowIndex < length) {
+        nums[slowIndex++] = 0;
+    }
+};
+```
+
+
+
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0344.反转字符串.md

@@ -190,13 +190,13 @@ javaScript:
  * @return {void} Do not return anything, modify s in-place instead.
  */
 var reverseString = function(s) {
-    return s.reverse();
+    //Do not return anything, modify s in-place instead.
+    reverse(s)
 };
 
-var reverseString = function(s) {
+var reverse = function(s) {
     let l = -1, r = s.length;
     while(++l < --r) [s[l], s[r]] = [s[r], s[l]];
-    return s;
 };
 ```
 
@@ -238,6 +238,22 @@ func reverseString(_ s: inout [Character]) {
 
 ```
 
+Rust:
+```Rust
+impl Solution {
+    pub fn reverse_string(s: &mut Vec<char>) {
+        let (mut left, mut right) = (0, s.len()-1);
+        while left < right {
+            let temp = s[left];
+            s[left] = s[right];
+            s[right] = temp;
+            left += 1;
+            right -= 1;
+        }
+    }
+}
+```
+
 C:
 ```c
 void reverseString(char* s, int sSize){

problems/0347.前K个高频元素.md

@@ -141,13 +141,10 @@ class Solution {
         }
 
         Set<Map.Entry<Integer, Integer>> entries = map.entrySet();
-        // 根据map的value值正序排，相当于一个小顶堆
-        PriorityQueue<Map.Entry<Integer, Integer>> queue = new PriorityQueue<>((o1, o2) -> o1.getValue() - o2.getValue());
+        // 根据map的value值，构建于一个大顶堆（o1 - o2: 小顶堆， o2 - o1 : 大顶堆）
+        PriorityQueue<Map.Entry<Integer, Integer>> queue = new PriorityQueue<>((o1, o2) -> o2.getValue() - o1.getValue());
         for (Map.Entry<Integer, Integer> entry : entries) {
             queue.offer(entry);
-            if (queue.size() > k) {
-                queue.poll();
-            }
         }
         for (int i = k - 1; i >= 0; i--) {
             result[i] = queue.poll().getKey();

problems/0349.两个数组的交集.md

@@ -137,13 +137,8 @@ class Solution {
                 resSet.add(i);
             }
         }
-        int[] resArr = new int[resSet.size()];
-        int index = 0;
         //将结果几何转为数组
-        for (int i : resSet) {
-            resArr[index++] = i;
-        }
-        return resArr;
+        return resSet.stream().mapToInt(x -> x).toArray();
     }
 }
 ```

problems/0376.摆动序列.md

@@ -228,6 +228,8 @@ class Solution {
 
 ### Python 
 
+**贪心**
+
 ```python
 class Solution:
     def wiggleMaxLength(self, nums: List[int]) -> int:
@@ -240,7 +242,30 @@ class Solution:
         return res
 ```
 
+**动态规划**
+
+```python
+class Solution:
+    def wiggleMaxLength(self, nums: List[int]) -> int:
+        # 0 i 作为波峰的最大长度
+        # 1 i 作为波谷的最大长度
+        # dp是一个列表，列表中每个元素是长度为 2 的列表
+        dp = []
+        for i in range(len(nums)):
+            # 初始为[1, 1]
+            dp.append([1, 1])
+            for j in range(i):
+                # nums[i] 为波谷
+                if nums[j] > nums[i]:
+                    dp[i][1] = max(dp[i][1], dp[j][0] + 1)
+                # nums[i] 为波峰 
+                if nums[j] < nums[i]:
+                    dp[i][0] = max(dp[i][0], dp[j][1] + 1)
+        return max(dp[-1][0], dp[-1][1])
+```
+
 ### Go 
+
 ```golang
 func wiggleMaxLength(nums []int) int {
     var count,preDiff,curDiff  int
@@ -298,9 +323,33 @@ var wiggleMaxLength = function(nums) {
 };
 ```
 
+### Rust
+**贪心**
+```Rust
+impl Solution {
+    pub fn wiggle_max_length(nums: Vec<i32>) -> i32 {
+        let len = nums.len() as usize;
+        if len <= 1 {
+            return len as i32;
+        }
+        let mut preDiff = 0;
+        let mut curDiff = 0;
+        let mut result = 1;
+        for i in 0..len-1 {
+            curDiff = nums[i+1] - nums[i];
+            if (preDiff <= 0 && curDiff > 0) || (preDiff >= 0 && curDiff < 0) {
+                result += 1;
+                preDiff = curDiff;
+            }
+        }
+        result
+    }
+}
+```
 
 ### C
 **贪心**
+
 ```c
 int wiggleMaxLength(int* nums, int numsSize){
     if(numsSize <= 1) 
@@ -326,6 +375,44 @@ int wiggleMaxLength(int* nums, int numsSize){
 }
 ```
 
+**动态规划**
+
+```c
+int max(int left, int right)
+{
+    return left > right ? left : right;
+}
+int wiggleMaxLength(int* nums, int numsSize){
+    if(numsSize <= 1)
+    {
+        return numsSize;
+    }
+    // 0   i 作为波峰的最大长度
+    // 1   i 作为波谷的最大长度
+    int dp[numsSize][2];
+    for(int i = 0; i < numsSize; i++)
+    {
+        dp[i][0] = 1;
+        dp[i][1] = 1;
+        for(int j = 0; j < i; j++)
+        {
+            // nums[i] 为山谷
+            if(nums[j] > nums[i])
+            {
+                dp[i][1] = max(dp[i][1], dp[j][0] + 1);
+            }
+            // nums[i] 为山峰
+            if(nums[j] < nums[i])
+            {
+                dp[i][0] = max(dp[i][0], dp[j][1] + 1);
+            }
+        }
+    }
+    return max(dp[numsSize - 1][0], dp[numsSize - 1][1]);
+}
+```
+
+
 
 ### TypeScript
 
@@ -375,7 +462,31 @@ function wiggleMaxLength(nums: number[]): number {
 };
 ```
 
+### Scala
 
+```scala
+object Solution {
+  def wiggleMaxLength(nums: Array[Int]): Int = {
+    if (nums.length <= 1) return nums.length
+    var result = 1
+    var curDiff = 0 // 当前一对的差值
+    var preDiff = 0 // 前一对的差值
+
+    for (i <- 1 until nums.length) {
+      curDiff = nums(i) - nums(i - 1) // 计算当前这一对的差值
+      // 当 curDiff > 0 的情况，preDiff <= 0
+      // 当 curDiff < 0 的情况，preDiff >= 0
+      // 这两种情况算是两个峰值
+      if ((curDiff > 0 && preDiff <= 0) || (curDiff < 0 && preDiff >= 0)) {
+        result += 1 // 结果集加 1
+        preDiff = curDiff // 当前差值赋值给上一轮
+      }
+    }
+
+    result
+  }
+}
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0450.删除二叉搜索树中的节点.md

@@ -456,31 +456,42 @@ func deleteNode(root *TreeNode, key int) *TreeNode {
  * @param {number} key
  * @return {TreeNode}
  */
-var deleteNode = function (root, key) {
-    if (root === null)
-        return root;
-    if (root.val === key) {
-        if (!root.left)
-            return root.right;
-        else if (!root.right)
-            return root.left;
-        else {
-            let cur = root.right;
-            while (cur.left) {
-                cur = cur.left;
-            }
-            cur.left = root.left;
-            root = root.right;
-            delete root;
-            return root;
-        }
-    }
-    if (root.val > key)
-        root.left = deleteNode(root.left, key);
-    if (root.val < key)
+var deleteNode = function(root, key) {
+    if (!root) return null;
+    if (key > root.val) {
         root.right = deleteNode(root.right, key);
-    return root;
+        return root;
+    } else if (key < root.val) {
+        root.left = deleteNode(root.left, key);
+        return root;
+    } else {
+        // 场景1: 该节点是叶节点
+        if (!root.left && !root.right) {
+            return null
+        }
+        // 场景2: 有一个孩子节点不存在
+        if (root.left && !root.right) {
+            return root.left;
+        } else if (root.right && !root.left) {
+            return root.right;
+        }
+        // 场景3: 左右节点都存在
+        const rightNode = root.right;
+        // 获取最小值节点
+        const minNode = getMinNode(rightNode);
+        // 将待删除节点的值替换为最小值节点值
+        root.val = minNode.val;
+        // 删除最小值节点
+        root.right = deleteNode(root.right, minNode.val);
+        return root;
+    }
 };
+function getMinNode(root) {
+    while (root.left) {
+        root = root.left;
+    }
+    return root;
+}
 ```
 
 迭代
@@ -582,7 +593,35 @@ function deleteNode(root: TreeNode | null, key: number): TreeNode | null {
 };
 ```
 
+## Scala
 
+```scala
+object Solution {
+  def deleteNode(root: TreeNode, key: Int): TreeNode = {
+    if (root == null) return root // 第一种情况，没找到删除的节点，遍历到空节点直接返回
+    if (root.value == key) {
+      // 第二种情况: 左右孩子都为空，直接删除节点，返回null
+      if (root.left == null && root.right == null) return null
+      // 第三种情况: 左孩子为空，右孩子不为空，右孩子补位
+      else if (root.left == null && root.right != null) return root.right
+      // 第四种情况: 左孩子不为空，右孩子为空，左孩子补位
+      else if (root.left != null && root.right == null) return root.left
+      // 第五种情况: 左右孩子都不为空，将删除节点的左子树头节点(左孩子)放到
+      //           右子树的最左边节点的左孩子上，返回删除节点的右孩子
+      else {
+        var tmp = root.right
+        while (tmp.left != null) tmp = tmp.left
+        tmp.left = root.left
+        return root.right
+      }
+    }
+    if (root.value > key) root.left = deleteNode(root.left, key)
+    if (root.value < key) root.right = deleteNode(root.right, key)
+
+    root // 返回根节点，return关键字可以省略
+  }
+}
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0452.用最少数量的箭引爆气球.md

@@ -136,17 +136,28 @@ public:
 
 ### Java 
 ```java
+/**
+时间复杂度 : O(NlogN)  排序需要 O(NlogN) 的复杂度
+
+空间复杂度 : O(logN) java所使用的内置函数用的是快速排序需要 logN 的空间
+*/
 class Solution {
     public int findMinArrowShots(int[][] points) {
         if (points.length == 0) return 0;
-        Arrays.sort(points, (o1, o2) -> Integer.compare(o1[0], o2[0]));
-
+        //用x[0] - y[0] 会大于2147483647 造成整型溢出
+        Arrays.sort(points, (x, y) -> Integer.compare(x[0], y[0]));
+        //count = 1 因为最少需要一个箭来射击第一个气球
         int count = 1;
-        for (int i = 1; i < points.length; i++) {
-            if (points[i][0] > points[i - 1][1]) {
+        //重叠气球的最小右边界
+        int leftmostRightBound  = points[0][1];
+        //如果下一个气球的左边界大于最小右边界
+            if (points[i][0] > leftmostRightBound ) {
+                //增加一次射击
                 count++;
+                leftmostRightBound  = points[i][1];
+                //不然就更新最小右边界
             } else {
-                points[i][1] = Math.min(points[i][1],points[i - 1][1]);
+                leftmostRightBound  = Math.min(leftmostRightBound , points[i][1]);
             }
         }
         return count;

problems/0455.分发饼干.md

@@ -296,5 +296,26 @@ int findContentChildren(int* g, int gSize, int* s, int sSize){
 }
 ```
 
+### Scala
+
+```scala
+object Solution {
+  def findContentChildren(g: Array[Int], s: Array[Int]): Int = {
+    var result = 0
+    var children = g.sorted
+    var cookie = s.sorted
+    // 遍历饼干
+    var j = 0
+    for (i <- cookie.indices) {
+      if (j < children.size && cookie(i) >= children(j)) {
+        j += 1
+        result += 1
+      }
+    }
+    result
+  }
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0459.重复的子字符串.md

@@ -505,5 +505,37 @@ Swift:
     }
 ```
 
+Rust:
+
+>前缀表统一不减一
+```Rust
+impl Solution {
+    pub fn get_next(next: &mut Vec<usize>, s: &Vec<char>) {
+        let len = s.len();
+        let mut j = 0;
+        for i in 1..len {
+            while j > 0 && s[i] != s[j] {
+                j = next[j - 1];
+            }
+            if s[i] == s[j] {
+                j += 1;
+            }
+            next[i] = j;
+        }
+    }
+
+    pub fn repeated_substring_pattern(s: String) -> bool {
+        let s = s.chars().collect::<Vec<char>>();
+        let len = s.len();
+        if len == 0 { return false; };
+        let mut next = vec![0; len];
+        Self::get_next(&mut next, &s);
+        if next[len - 1] != 0 && len % (len - (next[len - 1] )) == 0 { return true; }
+        return false;
+    }
+}
+```
+
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0491.递增子序列.md

@@ -522,5 +522,39 @@ func findSubsequences(_ nums: [Int]) -> [[Int]] {
 ```
 
 
+## Scala
+
+```scala
+object Solution {
+  import scala.collection.mutable
+  def findSubsequences(nums: Array[Int]): List[List[Int]] = {
+    var result = mutable.ListBuffer[List[Int]]()
+    var path = mutable.ListBuffer[Int]()
+
+    def backtracking(startIndex: Int): Unit = {
+      // 集合元素大于1，添加到结果集
+      if (path.size > 1) {
+        result.append(path.toList)
+      }
+
+      var used = new Array[Boolean](201)
+      // 使用循环守卫，当前层没有用过的元素才有资格进入回溯
+      for (i <- startIndex until nums.size if !used(nums(i) + 100)) {
+        // 如果path没元素或 当前循环的元素比path的最后一个元素大，则可以进入回溯
+        if (path.size == 0 || (!path.isEmpty && nums(i) >= path(path.size - 1))) {
+          used(nums(i) + 100) = true
+          path.append(nums(i))
+          backtracking(i + 1)
+          path.remove(path.size - 1)
+        }
+      }
+    }
+
+    backtracking(0)
+    result.toList
+  }
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0501.二叉搜索树中的众数.md

@@ -9,7 +9,9 @@
 
 # 501.二叉搜索树中的众数
 
-[力扣题目链接](https://leetcode.cn/problems/find-mode-in-binary-search-tree/solution/)
+
+[力扣题目链接](https://leetcode.cn/problems/find-mode-in-binary-search-tree/)
+
 
 给定一个有相同值的二叉搜索树（BST），找出 BST 中的所有众数（出现频率最高的元素）。
 
@@ -798,7 +800,76 @@ function findMode(root: TreeNode | null): number[] {
 };
 ```
 
+## Scala
 
+暴力:
+```scala
+object Solution {
+  // 导包
+  import scala.collection.mutable // 集合包
+  import scala.util.control.Breaks.{break, breakable} // 流程控制包
+  def findMode(root: TreeNode): Array[Int] = {
+    var map = mutable.HashMap[Int, Int]() // 存储节点的值，和该值出现的次数
+    def searchBST(curNode: TreeNode): Unit = {
+      if (curNode == null) return
+      var value = map.getOrElse(curNode.value, 0)
+      map.put(curNode.value, value + 1) 
+      searchBST(curNode.left)
+      searchBST(curNode.right)
+    }
+    searchBST(root) // 前序遍历把每个节点的值加入到里面
+    // 将map转换为list，随后根据元组的第二个值进行排序
+    val list = map.toList.sortWith((map1, map2) => {
+      if (map1._2 > map2._2) true else false
+    })
+    var res = mutable.ArrayBuffer[Int]()
+    res.append(list(0)._1) // 将第一个加入结果集
+    breakable {
+      for (i <- 1 until list.size) {
+        // 如果值相同就加入结果集合，反之break
+        if (list(i)._2 == list(0)._2) res.append(list(i)._1)
+        else break
+      }
+    }
+    res.toArray // 最终返回res的Array格式，return关键字可以省略
+  }
+}
+```
+
+递归(利用二叉搜索树的性质):
+```scala
+object Solution {
+  import scala.collection.mutable
+  def findMode(root: TreeNode): Array[Int] = {
+    var maxCount = 0 // 最大频率
+    var count = 0 // 统计频率
+    var pre: TreeNode = null
+    var result = mutable.ArrayBuffer[Int]()
+
+    def searchBST(cur: TreeNode): Unit = {
+      if (cur == null) return
+      searchBST(cur.left)
+      if (pre == null) count = 1 // 等于空置为1
+      else if (pre.value == cur.value) count += 1 // 与上一个节点的值相同加1
+      else count = 1 // 与上一个节点的值不同
+      pre = cur
+
+      // 如果和最大值相同，则放入结果集
+      if (count == maxCount) result.append(cur.value)
+
+      // 如果当前计数大于最大值频率，更新最大值，清空结果集
+      if (count > maxCount) {
+        maxCount = count
+        result.clear()
+        result.append(cur.value)
+      }
+      searchBST(cur.right)
+    }
+    searchBST(root)
+    result.toArray  // return关键字可以省略
+  }
+}
+```
 
 
 -----------------------

problems/0509.斐波那契数.md

@@ -234,6 +234,7 @@ func fib(n int) int {
 }
 ```
 ### Javascript 
+解法一
 ```Javascript
 var fib = function(n) {
     let dp = [0, 1]
@@ -244,6 +245,23 @@ var fib = function(n) {
     return dp[n]
 };
 ```
+解法二：时间复杂度O(N)，空间复杂度O(1)
+```Javascript
+var fib = function(n) {
+    // 动规状态转移中，当前结果只依赖前两个元素的结果，所以只要两个变量代替dp数组记录状态过程。将空间复杂度降到O(1)
+    let pre1 = 1
+    let pre2 = 0
+    let temp
+    if (n === 0) return 0
+    if (n === 1) return 1
+    for(let i = 2; i <= n; i++) {
+        temp = pre1
+        pre1 = pre1 + pre2
+        pre2 = temp
+    }
+    return pre1
+};
+```
 
 TypeScript
 

problems/0530.二叉搜索树的最小绝对差.md

@@ -431,7 +431,84 @@ function getMinimumDifference(root: TreeNode | null): number {
 };
 ```
 
+## Scala
 
+构建二叉树的有序数组:
+
+```scala
+object Solution {
+  import scala.collection.mutable
+  def getMinimumDifference(root: TreeNode): Int = {
+    val arr = mutable.ArrayBuffer[Int]()
+    def traversal(node: TreeNode): Unit = {
+      if (node == null) return
+      traversal(node.left)
+      arr.append(node.value)
+      traversal(node.right)
+    }
+    traversal(root)
+    // 在有序数组上求最小差值
+    var result = Int.MaxValue
+    for (i <- 1 until arr.size) {
+      result = math.min(result, arr(i) - arr(i - 1))
+    }
+    result // 返回最小差值
+  }
+}
+```
+
+递归记录前一个节点:
+
+```scala
+object Solution {
+  def getMinimumDifference(root: TreeNode): Int = {
+    var result = Int.MaxValue // 初始化为最大值
+    var pre: TreeNode = null // 记录前一个节点
+
+    def traversal(cur: TreeNode): Unit = {
+      if (cur == null) return
+      traversal(cur.left)
+      if (pre != null) {
+        // 对比result与节点之间的差值
+        result = math.min(result, cur.value - pre.value)
+      }
+      pre = cur
+      traversal(cur.right)
+    }
+
+    traversal(root)
+    result // return关键字可以省略
+  }
+}
+```
+
+迭代解决:
+
+```scala
+object Solution {
+  import scala.collection.mutable
+  def getMinimumDifference(root: TreeNode): Int = {
+    var result = Int.MaxValue // 初始化为最大值
+    var pre: TreeNode = null // 记录前一个节点
+    var cur = root
+    var stack = mutable.Stack[TreeNode]()
+    while (cur != null || !stack.isEmpty) {
+      if (cur != null) {
+        stack.push(cur)
+        cur = cur.left
+      } else {
+        cur = stack.pop()
+        if (pre != null) {
+          result = math.min(result, cur.value - pre.value)
+        }
+        pre = cur
+        cur = cur.right
+      }
+    }
+    result // return关键字可以省略
+  }
+}
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0538.把二叉搜索树转换为累加树.md

@@ -352,6 +352,24 @@ function convertBST(root: TreeNode | null): TreeNode | null {
 };
 ```
 
+## Scala
+
+```scala
+object Solution {
+  def convertBST(root: TreeNode): TreeNode = {
+    var sum = 0
+    def convert(node: TreeNode): Unit = {
+      if (node == null) return
+      convert(node.right)
+      sum += node.value
+      node.value = sum
+      convert(node.left)
+    }
+    convert(root)
+    root
+  }
+}
+```
 
 
 -----------------------

problems/0541.反转字符串II.md

@@ -53,10 +53,10 @@ public:
             // 2. 剩余字符小于 2k 但大于或等于 k 个，则反转前 k 个字符
             if (i + k <= s.size()) {
                 reverse(s.begin() + i, s.begin() + i + k );
-                continue;
+            } else {
+                // 3. 剩余字符少于 k 个，则将剩余字符全部反转。
+                reverse(s.begin() + i, s.end());
             }
-            // 3. 剩余字符少于 k 个，则将剩余字符全部反转。
-            reverse(s.begin() + i, s.begin() + s.size());
         }
         return s;
     }
@@ -389,5 +389,36 @@ object Solution {
   }
 }
 ```
+
+Rust:
+
+```Rust
+impl Solution {
+    pub fn reverse(s: &mut Vec<char>, mut begin: usize, mut end: usize){
+        while begin < end {
+            let temp = s[begin];
+            s[begin] = s[end];
+            s[end] = temp;
+            begin += 1;
+            end -= 1;
+        }
+    }
+    pub fn reverse_str(s: String, k: i32) -> String {
+        let len = s.len();
+        let k = k as usize;
+        let mut s = s.chars().collect::<Vec<_>>();
+        for i in (0..len).step_by(2 * k) {
+            if i + k < len {
+                Self::reverse(&mut s, i, i + k - 1);
+            }
+            else {
+                Self::reverse(&mut s, i, len - 1);
+            }
+        }
+        s.iter().collect::<String>()
+    }
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0617.合并二叉树.md

@@ -631,7 +631,60 @@ function mergeTrees(root1: TreeNode | null, root2: TreeNode | null): TreeNode |
 };
 ```
 
+## Scala
 
+递归:
+```scala
+object Solution {
+  def mergeTrees(root1: TreeNode, root2: TreeNode): TreeNode = {
+    if (root1 == null) return root2 // 如果root1为空，返回root2
+    if (root2 == null) return root1 // 如果root2为空，返回root1
+    // 新建一个节点，值为两个节点的和
+    var node = new TreeNode(root1.value + root2.value)
+    // 往下递归
+    node.left = mergeTrees(root1.left, root2.left)
+    node.right = mergeTrees(root1.right, root2.right)
+    node // 返回node，return关键字可以省略
+  }
+}
+```
+
+迭代:
+```scala
+object Solution {
+  import scala.collection.mutable
+  def mergeTrees(root1: TreeNode, root2: TreeNode): TreeNode = {
+    if (root1 == null) return root2
+    if (root2 == null) return root1
+    var stack = mutable.Stack[TreeNode]()
+    // 先放node2再放node1
+    stack.push(root2)
+    stack.push(root1)
+    while (!stack.isEmpty) {
+      var node1 = stack.pop()
+      var node2 = stack.pop()
+      node1.value += node2.value
+      if (node1.right != null && node2.right != null) {
+        stack.push(node2.right)
+        stack.push(node1.right)
+      } else {
+        if(node1.right == null){
+          node1.right = node2.right
+        }
+      }
+      if (node1.left != null && node2.left != null) {
+        stack.push(node2.left)
+        stack.push(node1.left)
+      } else {
+        if(node1.left == null){
+          node1.left = node2.left
+        }
+      }
+    }
+    root1
+  }
+}
+```
 
 
 -----------------------

problems/0654.最大二叉树.md

@@ -476,7 +476,33 @@ func traversal(_ nums: inout [Int], _ left: Int, _ right: Int) -> TreeNode? {
 }
 ```
 
+## Scala
 
+```scala
+object Solution {
+  def constructMaximumBinaryTree(nums: Array[Int]): TreeNode = {
+    if (nums.size == 0) return null
+    // 找到数组最大值
+    var maxIndex = 0
+    var maxValue = Int.MinValue
+    for (i <- nums.indices) {
+      if (nums(i) > maxValue) {
+        maxIndex = i
+        maxValue = nums(i)
+      }
+    }
+
+    // 构建一棵树
+    var root = new TreeNode(maxValue, null, null)
+
+    // 递归寻找左右子树
+    root.left = constructMaximumBinaryTree(nums.slice(0, maxIndex))
+    root.right = constructMaximumBinaryTree(nums.slice(maxIndex + 1, nums.length))
+
+    root // 返回root
+  }
+}
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0669.修剪二叉搜索树.md

@@ -453,7 +453,21 @@ function trimBST(root: TreeNode | null, low: number, high: number): TreeNode | n
 };
 ```
 
+## Scala
 
+递归法:
+```scala
+object Solution {
+  def trimBST(root: TreeNode, low: Int, high: Int): TreeNode = {
+    if (root == null) return null
+    if (root.value < low) return trimBST(root.right, low, high)
+    if (root.value > high) return trimBST(root.left, low, high)
+    root.left = trimBST(root.left, low, high)
+    root.right = trimBST(root.right, low, high)
+    root
+  }
+}
+```
 
 
 -----------------------

problems/0700.二叉搜索树中的搜索.md

@@ -363,7 +363,34 @@ function searchBST(root: TreeNode | null, val: number): TreeNode | null {
 };
 ```
 
+## Scala
 
+递归:
+```scala
+object Solution {
+  def searchBST(root: TreeNode, value: Int): TreeNode = {
+    if (root == null || value == root.value) return root
+    // 相当于三元表达式，在Scala中if...else有返回值
+    if (value < root.value) searchBST(root.left, value) else searchBST(root.right, value)
+  }
+}
+```
+
+迭代:
+```scala
+object Solution {
+  def searchBST(root: TreeNode, value: Int): TreeNode = {
+    // 因为root是不可变量，所以需要赋值给一个可变量
+    var node = root
+    while (node != null) {
+      if (value < node.value) node = node.left
+      else if (value > node.value) node = node.right
+      else return node
+    }
+    null // 没有返回就返回空
+  }
+}
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0701.二叉搜索树中的插入操作.md

@@ -585,6 +585,43 @@ function insertIntoBST(root: TreeNode | null, val: number): TreeNode | null {
 ```
 
 
+## Scala
+
+递归:
+
+```scala
+object Solution {
+  def insertIntoBST(root: TreeNode, `val`: Int): TreeNode = {
+    if (root == null) return new TreeNode(`val`)
+    if (`val` < root.value) root.left = insertIntoBST(root.left, `val`) 
+    else root.right = insertIntoBST(root.right, `val`)
+    root  // 返回根节点
+  }
+}
+```
+
+迭代:
+
+```scala
+object Solution {
+  def insertIntoBST(root: TreeNode, `val`: Int): TreeNode = {
+    if (root == null) {
+      return new TreeNode(`val`)
+    }
+    var parent = root   // 记录当前节点的父节点
+    var curNode = root
+    while (curNode != null) {
+      parent = curNode
+      if(`val` < curNode.value) curNode = curNode.left
+      else curNode = curNode.right
+    }
+    if(`val` < parent.value) parent.left = new TreeNode(`val`)
+    else parent.right = new TreeNode(`val`)
+    root  // 最终返回根节点
+  }
+}
+```
+
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0704.二分查找.md

@@ -613,6 +613,36 @@ public class Solution{
 }
 ```
 
+**Kotlin:**
+```kotlin
+class Solution {
+    fun search(nums: IntArray, target: Int): Int {
+	// leftBorder
+        var left:Int = 0
+	// rightBorder
+        var right:Int = nums.size - 1
+	// 使用左闭右闭区间
+        while (left <= right) {
+            var middle:Int = left + (right - left)/2
+	    // taget 在左边
+            if (nums[middle] > target) {
+                right = middle - 1
+            }
+            else {
+		// target 在右边
+                if (nums[middle] < target) {
+                    left = middle + 1
+                }
+		// 找到了，返回
+                else   return middle
+                }
+            }
+	    // 没找到，返回
+            return -1   
+        }
+}
+```
+
 
 
 **Kotlin:**

problems/0707.设计链表.md

@@ -106,8 +106,9 @@ public:
     // 在第index个节点之前插入一个新节点，例如index为0，那么新插入的节点为链表的新头节点。
     // 如果index 等于链表的长度，则说明是新插入的节点为链表的尾结点
     // 如果index大于链表的长度，则返回空
+    // 如果index小于0，则置为0，作为链表的新头节点。
     void addAtIndex(int index, int val) {
-        if (index > _size) {
+        if (index > _size || index < 0) {
             return;
         }
         LinkedNode* newNode = new LinkedNode(val);

problems/0724.寻找数组的中心索引.md

@@ -140,6 +140,24 @@ var pivotIndex = function(nums) {
 };
 ```
 
+### TypeScript
+
+```typescript
+function pivotIndex(nums: number[]): number {
+    const length: number = nums.length;
+    const sum: number = nums.reduce((a, b) => a + b);
+    let leftSum: number = 0;
+    for (let i = 0; i < length; i++) {
+        const rightSum: number = sum - leftSum - nums[i];
+        if (leftSum === rightSum) return i;
+        leftSum += nums[i];
+    }
+    return -1;
+};
+```
+
+
+
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0746.使用最小花费爬楼梯.md

@@ -288,6 +288,24 @@ function minCostClimbingStairs(cost: number[]): number {
 };
 ```
 
+### Rust
+
+```Rust
+use std::cmp::min;
+impl Solution {
+    pub fn min_cost_climbing_stairs(cost: Vec<i32>) -> i32 {
+        let len = cost.len();
+        let mut dp = vec![0; len];
+        dp[0] = cost[0];
+        dp[1] = cost[1];
+        for i in 2..len {
+            dp[i] = min(dp[i-1], dp[i-2]) + cost[i];
+        }
+        min(dp[len-1], dp[len-2])
+    }
+}
+```
+
 ### C
 
 ```c

problems/0844.比较含退格的字符串.md

@@ -399,6 +399,71 @@ var backspaceCompare = function(s, t) {
 
 ```
 
+### TypeScript
+
+> 双栈法：
+
+```typescript
+function backspaceCompare(s: string, t: string): boolean {
+    const stack1: string[] = [],
+        stack2: string[] = [];
+    for (let c of s) {
+        if (c === '#') {
+            stack1.pop();
+        } else {
+            stack1.push(c);
+        }
+    }
+    for (let c of t) {
+        if (c === '#') {
+            stack2.pop();
+        } else {
+            stack2.push(c);
+        }
+    }
+    if (stack1.length !== stack2.length) return false;
+    for (let i = 0, length = stack1.length; i < length; i++) {
+        if (stack1[i] !== stack2[i]) return false;
+    }
+    return true;
+};
+```
+
+> 双指针法：
+
+```typescript
+function backspaceCompare(s: string, t: string): boolean {
+    let sIndex: number = s.length - 1,
+        tIndex: number = t.length - 1;
+    while (true) {
+        sIndex = getIndexAfterDel(s, sIndex);
+        tIndex = getIndexAfterDel(t, tIndex);
+        if (sIndex < 0 || tIndex < 0) break;
+        if (s[sIndex] !== t[tIndex]) return false;
+        sIndex--;
+        tIndex--;
+    }
+    return sIndex === -1 && tIndex === -1;
+};
+function getIndexAfterDel(s: string, startIndex: number): number {
+    let backspaceNum: number = 0;
+    while (startIndex >= 0) {
+        // 不可消除
+        if (s[startIndex] !== '#' && backspaceNum === 0) break;
+        // 可消除
+        if (s[startIndex] === '#') {
+            backspaceNum++;
+        } else {
+            backspaceNum--;
+        }
+        startIndex--;
+    }
+    return startIndex;
+}
+```
+
+
+
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0922.按奇偶排序数组II.md

@@ -260,6 +260,75 @@ var sortArrayByParityII = function(nums) {
 };
 ```
 
+### TypeScript
+
+> 方法一：
+
+```typescript
+function sortArrayByParityII(nums: number[]): number[] {
+    const evenArr: number[] = [],
+        oddArr: number[] = [];
+    for (let num of nums) {
+        if (num % 2 === 0) {
+            evenArr.push(num);
+        } else {
+            oddArr.push(num);
+        }
+    }
+    const resArr: number[] = [];
+    for (let i = 0, length = nums.length / 2; i < length; i++) {
+        resArr.push(evenArr[i]);
+        resArr.push(oddArr[i]);
+    }
+    return resArr;
+};
+```
+
+> 方法二：
+
+```typescript
+function sortArrayByParityII(nums: number[]): number[] {
+    const length: number = nums.length;
+    const resArr: number[] = [];
+    let evenIndex: number = 0,
+        oddIndex: number = 1;
+    for (let i = 0; i < length; i++) {
+        if (nums[i] % 2 === 0) {
+            resArr[evenIndex] = nums[i];
+            evenIndex += 2;
+        } else {
+            resArr[oddIndex] = nums[i];
+            oddIndex += 2;
+        }
+    }
+    return resArr;
+};
+```
+
+> 方法三：
+
+```typescript
+function sortArrayByParityII(nums: number[]): number[] {
+    const length: number = nums.length;
+    let oddIndex: number = 1;
+    for (let evenIndex = 0; evenIndex < length; evenIndex += 2) {
+        if (nums[evenIndex] % 2 === 1) {
+            // 在偶数位遇到了奇数
+            while (oddIndex < length && nums[oddIndex] % 2 === 1) {
+                oddIndex += 2;
+            }
+            // 在奇数位遇到了偶数，交换
+            let temp = nums[evenIndex];
+            nums[evenIndex] = nums[oddIndex];
+            nums[oddIndex] = temp;
+        }
+    }
+    return nums;
+};
+```
+
+
+
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0925.长按键入.md

@@ -209,6 +209,31 @@ var isLongPressedName = function(name, typed) {
 };
 ```
 
+### TypeScript
+
+```typescript
+function isLongPressedName(name: string, typed: string): boolean {
+    const nameLength: number = name.length,
+        typeLength: number = typed.length;
+    let i: number = 0,
+        j: number = 0;
+    while (i < nameLength && j < typeLength) {
+        if (name[i] !== typed[j]) return false;
+        i++;
+        j++;
+        if (i === nameLength || name[i] !== name[i - 1]) {
+            // 跳过typed中的连续相同字符
+            while (j < typeLength && typed[j] === typed[j - 1]) {
+                j++;
+            }
+        }
+    }
+    return i === nameLength && j === typeLength;
+};
+```
+
+
+
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0977.有序数组的平方.md

@@ -106,6 +106,7 @@ class Solution {
         int index = result.length - 1;
         while (left <= right) {
             if (nums[left] * nums[left] > nums[right] * nums[right]) {
+                // 正数的相对位置是不变的， 需要调整的是负数平方后的相对位置
                 result[index--] = nums[left] * nums[left];
                 ++left;
             } else {
@@ -362,6 +363,8 @@ class Solution {
 ```
 
 Kotlin:
+
+双指针法
 ```kotlin
 class Solution {
     // 双指针法
@@ -383,6 +386,32 @@ class Solution {
     }
 }
 ```
+骚操作（暴力思路）
+```kotlin
+class Solution {
+    fun sortedSquares(nums: IntArray): IntArray {
+        // left 与 right 用来控制循环，类似于滑动窗口
+        var left: Int = 0;
+        var right: Int = nums.size - 1;
+        // 将每个数字的平方经过排序后加入result数值
+        var result: IntArray = IntArray(nums.size);
+        var k: Int = nums.size - 1;
+        while (left <= right) {
+            // 从大到小，从后向前填满数组
+            // [left, right] 控制循环
+            if (nums[left] * nums[left] > nums[right] * nums[right]) {
+                result[k--] = nums[left] * nums[left]
+                left++
+            }
+            else {
+                result[k--] = nums[right] * nums[right]
+                right--
+            }
+        }
+        return result
+    }
+}
+```
 
 Scala:
 
@@ -420,6 +449,24 @@ object Solution {
 }
 ```
 
-
+C#：
+```csharp
+public class Solution {
+    public int[] SortedSquares(int[] nums) {
+        int k = nums.Length - 1;
+        int[] result = new int[nums.Length];
+        for (int i = 0, j = nums.Length - 1;i <= j;){
+            if (nums[i] * nums[i] < nums[j] * nums[j]) {
+                result[k--] = nums[j] * nums[j];
+                j--;
+            } else {
+                result[k--] = nums[i] * nums[i];
+                i++;
+            }
+        }
+        return result;
+    }
+}
+```
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/1047.删除字符串中的所有相邻重复项.md

@@ -374,6 +374,34 @@ func removeDuplicates(_ s: String) -> String {
     return String(stack)
 }
 ```
+
+
+PHP:
+```php
+class Solution {
+    function removeDuplicates($s) {
+        $stack = new SplStack();
+        for($i=0;$i<strlen($s);$i++){
+            if($stack->isEmpty() || $s[$i] != $stack->top()){
+                $stack->push($s[$i]);
+            }else{
+               $stack->pop(); 
+            }
+        }
+
+        $result = "";
+        while(!$stack->isEmpty()){
+            $result.= $stack->top();
+            $stack->pop();
+        }
+        
+        // 此时字符串需要反转一下
+        return strrev($result);
+    }
+}
+```
+
+
 Scala:
 ```scala
 object Solution {
@@ -396,5 +424,6 @@ object Solution {
   }
 }
 ```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/1049.最后一块石头的重量II.md

@@ -277,6 +277,37 @@ var lastStoneWeightII = function (stones) {
 };
 ```
 
+C版本
+```c
+#define MAX(a, b) (((a) > (b)) ? (a) : (b))
+
+int getSum(int *stones, int stoneSize) {
+    int sum = 0, i;
+    for (i = 0; i < stoneSize; ++i)
+        sum += stones[i];
+    return sum;
+}
+
+int lastStoneWeightII(int* stones, int stonesSize){
+    int sum = getSum(stones, stonesSize);
+    int target = sum / 2;
+    int i, j;
+
+    // 初始化dp数组
+    int *dp = (int*)malloc(sizeof(int) * (target + 1));
+    memset(dp, 0, sizeof(int) * (target + 1));
+    for (j = stones[0]; j <= target; ++j)
+        dp[j] = stones[0];
+    
+    // 递推公式：dp[j] = max(dp[j], dp[j - stones[i]] + stones[i])
+    for (i = 1; i < stonesSize; ++i) {
+        for (j = target; j >= stones[i]; --j)
+            dp[j] = MAX(dp[j], dp[j - stones[i]] + stones[i]);
+    }
+    return sum - dp[target] - dp[target];
+}
+```
+
 
 TypeScript版本
 

problems/1143.最长公共子序列.md

@@ -129,6 +129,9 @@ public:
 Java：
 
 ```java
+/*
+	二维dp数组
+*/
 class Solution {
     public int longestCommonSubsequence(String text1, String text2) {
         int[][] dp = new int[text1.length() + 1][text2.length() + 1]; // 先对dp数组做初始化操作
@@ -146,6 +149,47 @@ class Solution {
         return dp[text1.length()][text2.length()];
     }
 }
+
+
+
+/**
+	一维dp数组
+*/
+class Solution {
+    public int longestCommonSubsequence(String text1, String text2) {
+        int n1 = text1.length();
+        int n2 = text2.length();
+
+        // 多从二维dp数组过程分析  
+        // 关键在于  如果记录  dp[i - 1][j - 1]
+        // 因为 dp[i - 1][j - 1]  <!=>  dp[j - 1]  <=>  dp[i][j - 1]
+        int [] dp = new int[n2 + 1];
+
+        for(int i = 1; i <= n1; i++){
+
+            // 这里pre相当于 dp[i - 1][j - 1]
+            int pre = dp[0];
+            for(int j = 1; j <= n2; j++){
+
+                //用于给pre赋值
+                int cur = dp[j];
+                if(text1.charAt(i - 1) == text2.charAt(j - 1)){
+                    //这里pre相当于dp[i - 1][j - 1]   千万不能用dp[j - 1] !!
+                    dp[j] = pre + 1;
+                } else{
+                    // dp[j]     相当于   dp[i - 1][j]
+                    // dp[j - 1] 相当于   dp[i][j - 1]
+                    dp[j] = Math.max(dp[j], dp[j - 1]);
+                }
+
+                //更新dp[i - 1][j - 1], 为下次使用做准备
+                pre = cur;
+            }
+        }
+
+        return dp[n2];
+    }
+}
 ```
 
 Python：

problems/1382.将二叉搜索树变平衡.md

@@ -123,6 +123,46 @@ class Solution:
 ```
 Go：
 
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func balanceBST(root *TreeNode) *TreeNode {
+   // 二叉搜索树中序遍历得到有序数组
+	nums := []int{}
+   // 中序递归遍历二叉树
+	var travel func(node *TreeNode)
+	travel = func(node *TreeNode) {
+		if node == nil {
+			return
+		}
+		travel(node.Left)
+		nums = append(nums, node.Val)
+		travel(node.Right)
+	}
+	// 二分法保证左右子树高度差不超过一（题目要求返回的仍是二叉搜索树）
+	var buildTree func(nums []int, left, right int) *TreeNode
+	buildTree = func(nums []int, left, right int) *TreeNode {
+		if left > right {
+			return nil
+		}
+		mid := left + (right-left) >> 1
+		root := &TreeNode{Val: nums[mid]}
+		root.Left = buildTree(nums, left, mid-1)
+		root.Right = buildTree(nums, mid+1, right)
+		return root
+	}
+	travel(root)
+	return buildTree(nums, 0, len(nums)-1)
+}
+
+```
+
 JavaScript：
 ```javascript
 var balanceBST = function(root) {
@@ -148,6 +188,30 @@ var balanceBST = function(root) {
 };
 ```
 
+TypeScript：
+
+```typescript
+function balanceBST(root: TreeNode | null): TreeNode | null {
+    const inorderArr: number[] = [];
+    inorderTraverse(root, inorderArr);
+    return buildTree(inorderArr, 0, inorderArr.length - 1);
+};
+function inorderTraverse(node: TreeNode | null, arr: number[]): void {
+    if (node === null) return;
+    inorderTraverse(node.left, arr);
+    arr.push(node.val);
+    inorderTraverse(node.right, arr);
+}
+function buildTree(arr: number[], left: number, right: number): TreeNode | null {
+    if (left > right) return null;
+    const mid = (left + right) >> 1;
+    const resNode: TreeNode = new TreeNode(arr[mid]);
+    resNode.left = buildTree(arr, left, mid - 1);
+    resNode.right = buildTree(arr, mid + 1, right);
+    return resNode;
+}
+```
+
 
 
 -----------------------

problems/剑指Offer05.替换空格.md

@@ -506,6 +506,37 @@ function spaceLen($s){
 }
 ```
 
+Rust
+
+```Rust
+impl Solution {
+    pub fn replace_space(s: String) -> String {
+        let mut len: usize = s.len();
+        let mut s = s.chars().collect::<Vec<char>>();
+        let mut count = 0;
+        for i in &s {
+            if i.is_ascii_whitespace() {
+                count += 1;
+            }
+        }
+        let mut new_len = len + count * 2;
+        s.resize(new_len, ' ');
+        while len < new_len {
+            len -= 1;
+            new_len -= 1;
+            if s[len].is_ascii_whitespace() {
+                s[new_len] = '0';
+                s[new_len - 1] = '2';
+                s[new_len - 2] = '%';
+                new_len -= 2;
+            }
+            else { s[new_len] = s[len] }
+        }
+        s.iter().collect::<String>()
+    }
+}
+```
+
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/剑指Offer58-II.左旋转字符串.md

@@ -341,7 +341,30 @@ object Solution {
 }
 ```
 
+Rust:
 
+```Rust
+impl Solution {
+    pub fn reverse(s: &mut Vec<char>, mut begin: usize, mut end: usize){
+        while begin < end {
+            let temp = s[begin];
+            s[begin] = s[end];
+            s[end] = temp;
+            begin += 1;
+            end -= 1;
+        }
+    }
+    pub fn reverse_left_words(s: String, n: i32) -> String {
+        let len = s.len();
+        let mut s = s.chars().collect::<Vec<char>>();
+        let n = n as usize;
+        Self::reverse(&mut s, 0, n - 1);
+        Self::reverse(&mut s, n, len - 1);
+        Self::reverse(&mut s, 0, len - 1);
+        s.iter().collect::<String>()
+    }
+}
+```
 
 
 

problems/背包理论基础01背包-1.md

@@ -356,9 +356,13 @@ func test_2_wei_bag_problem1(weight, value []int, bagweight int) int {
 	// 递推公式
 	for i := 1; i < len(weight); i++ {
 		//正序,也可以倒序
-		for  j := weight[i];j<= bagweight ; j++ {
-			dp[i][j] = max(dp[i-1][j], dp[i-1][j-weight[i]]+value[i])
-		}		
+	    for j := 0; j <= bagweight; j++ {
+	        if j < weight[i] {
+	            dp[i][j] = dp[i-1][j]
+	        } else {
+	            dp[i][j] = max(dp[i-1][j], dp[i-1][j-weight[i]]+value[i])
+	        }
+	    }		
 	}
 	return dp[len(weight)-1][bagweight]
 }

problems/面试题02.07.链表相交.md

@@ -13,21 +13,21 @@
 
 图示两个链表在节点 c1 开始相交：
 
-![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221657.png) 
+![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221657.png)
 
 题目数据 保证 整个链式结构中不存在环。
 
-注意，函数返回结果后，链表必须 保持其原始结构 。 
+注意，函数返回结果后，链表必须 保持其原始结构 。
 
-示例 1： 
+示例 1：
 
-![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221723.png) 
+![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221723.png)
 
 示例 2：
 
-![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221749.png) 
+![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221749.png)
 
-示例 3： 
+示例 3：
 
 ![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221812.png)![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221812.png)
 
@@ -100,7 +100,7 @@ public:
 ## 其他语言版本
 
 
-### Java 
+### Java
 ```Java
 public class Solution {
     public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
@@ -144,11 +144,11 @@ public class Solution {
         }
         return null;
     }
-    
+
 }
 ```
 
-### Python 
+### Python
 ```python
 
 class Solution:
@@ -162,15 +162,15 @@ class Solution:
         """
         cur_a, cur_b = headA, headB     # 用两个指针代替a和b
 
-        
+
         while cur_a != cur_b:
             cur_a = cur_a.next if cur_a else headB      # 如果a走完了，那么就切换到b走
             cur_b = cur_b.next if cur_b else headA      # 同理，b走完了就切换到a
-        
+
         return cur_a
 ```
 
-### Go 
+### Go
 
 ```go
 func getIntersectionNode(headA, headB *ListNode) *ListNode {
@@ -208,7 +208,30 @@ func getIntersectionNode(headA, headB *ListNode) *ListNode {
 }
 ```
 
-### javaScript 
+双指针
+
+```go
+func getIntersectionNode(headA, headB *ListNode) *ListNode {
+    l1,l2 := headA, headB
+    for l1 != l2 {
+        if l1 != nil {
+            l1 = l1.Next
+        } else {
+            l1 = headB
+        }
+
+        if l2 != nil {
+            l2 = l2.Next
+        } else {
+            l2 = headA
+        }
+    }
+
+    return l1
+}
+```
+
+### javaScript
 
 ```js
 var getListLen = function(head) {
@@ -218,9 +241,9 @@ var getListLen = function(head) {
        cur = cur.next;
     }
     return len;
-} 
+}
 var getIntersectionNode = function(headA, headB) {
-    let curA = headA,curB = headB, 
+    let curA = headA,curB = headB,
         lenA = getListLen(headA),
         lenB = getListLen(headB);
     if(lenA < lenB) {