Merge branch 'master' into patch-45
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@@ -33,7 +33,7 @@
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## 算法公开课
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**《代码随想录》算法视频公开课:[动态规划之子序列问题,元素不连续!| LeetCode:300.最长递增子序列](https://www.bilibili.com/video/BV1ng411J7xP),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
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**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html)::[动态规划之子序列问题,元素不连续!| LeetCode:300.最长递增子序列](https://www.bilibili.com/video/BV1ng411J7xP),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
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## 思路
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@@ -124,8 +124,8 @@ public:
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## 其他语言版本
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### Java:
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Java:
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```Java
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class Solution {
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public int lengthOfLIS(int[] nums) {
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@@ -145,7 +145,7 @@ class Solution {
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}
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```
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Python:
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### Python:
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DP
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```python
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@@ -187,7 +187,8 @@ class Solution:
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return len(tails) # 返回递增子序列的长度
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```
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Go:
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### Go:
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```go
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// 动态规划求解
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func lengthOfLIS(nums []int) int {
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@@ -246,7 +247,8 @@ func lengthOfLIS(nums []int ) int {
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}
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```
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Javascript
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### Javascript:
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```javascript
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const lengthOfLIS = (nums) => {
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let dp = Array(nums.length).fill(1);
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@@ -265,7 +267,7 @@ const lengthOfLIS = (nums) => {
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};
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```
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TypeScript
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### TypeScript:
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```typescript
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function lengthOfLIS(nums: number[]): number {
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@@ -286,7 +288,8 @@ function lengthOfLIS(nums: number[]): number {
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};
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```
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Rust:
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### Rust:
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```rust
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pub fn length_of_lis(nums: Vec<i32>) -> i32 {
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let mut dp = vec![1; nums.len()];
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@@ -308,3 +311,4 @@ pub fn length_of_lis(nums: Vec<i32>) -> i32 {
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<a href="https://programmercarl.com/other/kstar.html" target="_blank">
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<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
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</a>
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