Update

Author: youngyangyang04

problems/0020.有效的括号.md

@@ -402,6 +402,38 @@ bool isValid(char * s){
 ```
 
 
+C#:
+```csharp
+public class Solution {
+    public bool IsValid(string s) {
+        var len = s.Length;
+        if(len % 2 == 1) return false; // 字符串长度为单数，直接返回 false
+        // 初始化栈
+        var stack = new Stack<char>();
+        // 遍历字符串
+        for(int i = 0; i < len; i++){
+            // 当字符串为左括号时，进栈对应的右括号
+            if(s[i] == '('){
+                stack.Push(')');
+            }else if(s[i] == '['){
+                stack.Push(']');
+            }else if(s[i] == '{'){
+                stack.Push('}');
+            }
+            // 当字符串为右括号时，当栈为空（无左括号） 或者 出栈字符不是当前的字符
+            else if(stack.Count == 0 || stack.Pop() != s[i])
+                return false;
+        }
+        // 如果栈不为空，例如“((()”，右括号少于左括号，返回false
+        if (stack.Count > 0)
+            return false;
+        // 上面的校验都满足，则返回true
+        else
+            return true;
+    }
+}
+```
+
 PHP:
 ```php
 // https://www.php.net/manual/zh/class.splstack.php

problems/0031.下一个排列.md

@@ -116,6 +116,48 @@ class Solution {
     }
 }
 ```
+> 优化时间复杂度为O(N)，空间复杂度为O(1)
+```Java
+class Solution {
+    public void nextPermutation(int[] nums) {
+        // 1.从后向前获取逆序区域的前一位
+        int index = findIndex(nums);
+        // 判断数组是否处于最小组合状态
+        if(index != 0){
+            // 2.交换逆序区域刚好大于它的最小数字
+            exchange(nums,index);
+        }
+        // 3.把原来的逆序区转为顺序
+        reverse(nums,index);
+    }
+    
+    public static int findIndex(int [] nums){
+        for(int i = nums.length-1;i>0;i--){
+            if(nums[i]>nums[i-1]){
+                return i;
+            }
+        }
+        return 0;
+    }
+    public static void exchange(int [] nums, int index){
+        int head = nums[index-1];
+        for(int i = nums.length-1;i>0;i--){
+            if(head < nums[i]){
+                nums[index-1] = nums[i];
+                nums[i] = head;
+                break;
+            }
+        }
+    }
+    public static void reverse(int [] nums, int index){
+        for(int i = index,j = nums.length-1;i<j;i++,j--){
+            int temp = nums[i];
+            nums[i] = nums[j];
+            nums[j] = temp;
+        }
+    }   
+}
+```
 
 ## Python
 >直接使用sorted()不符合题意
@@ -164,7 +206,7 @@ class Solution:
             high -= 1
 ```
 >上一版本简化版
-'''python
+```python
 class Solution(object):
     def nextPermutation(self, nums: List[int]) -> None:
         n = len(nums)
@@ -185,7 +227,7 @@ class Solution(object):
             end -= 1
         
         return nums
-'''
+```
 
 ## Go
 

problems/0035.搜索插入位置.md

@@ -142,7 +142,7 @@ public:
 ```
 
 * 时间复杂度：O(log n)
-* 时间复杂度：O(1) 
+* 空间复杂度：O(1)
 
 效率如下：
 ![35_搜索插入位置2](https://img-blog.csdnimg.cn/2020121623272877.png)

problems/0039.组合总和.md

@@ -417,6 +417,35 @@ function combinationSum(candidates: number[], target: number): number[][] {
 };
 ```
 
+## Rust
+
+```Rust
+impl Solution {
+    pub fn backtracking(result: &mut Vec<Vec<i32>>, path: &mut Vec<i32>, candidates: &Vec<i32>, target: i32, mut sum: i32, start_index: usize) {
+        if sum == target { 
+            result.push(path.to_vec());
+            return;
+        }
+        for i in start_index..candidates.len() {
+            if sum + candidates[i] <= target {
+                sum += candidates[i];
+                path.push(candidates[i]);
+                Self::backtracking(result, path, candidates, target, sum, i);
+                sum -= candidates[i];
+                path.pop();
+            }
+        }
+    }
+
+    pub fn combination_sum(candidates: Vec<i32>, target: 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, &candidates, target, 0, 0);
+        result
+    }
+}
+```
+
 ## C 
 
 ```c

problems/0040.组合总和II.md

@@ -599,6 +599,41 @@ function combinationSum2(candidates: number[], target: number): number[][] {
 };
 ```
 
+## Rust
+
+```Rust
+impl Solution {
+    pub fn backtracking(result: &mut Vec<Vec<i32>>, path: &mut Vec<i32>, candidates: &Vec<i32>, target: i32, mut sum: i32, start_index: usize, used: &mut Vec<bool>) {
+        if sum == target {
+            result.push(path.to_vec());
+            return;
+        }
+        for i in start_index..candidates.len() {
+            if sum + candidates[i] <= target {
+                if i > 0 && candidates[i] == candidates[i - 1] && used[i - 1] == false { continue; }
+                sum += candidates[i];
+                path.push(candidates[i]);
+                used[i] = true;
+                Self::backtracking(result, path, candidates, target, sum, i + 1, used);
+                used[i] = false;
+                sum -= candidates[i];
+                path.pop();
+            }
+        }
+    }
+
+    pub fn combination_sum2(candidates: Vec<i32>, target: i32) -> Vec<Vec<i32>> {
+        let mut result: Vec<Vec<i32>> = Vec::new();
+        let mut path: Vec<i32> = Vec::new();
+        let mut used: Vec<bool> = vec![false; candidates.len()];
+        let mut candidates = candidates;
+        candidates.sort();
+        Self::backtracking(&mut result, &mut path, &candidates, target, 0, 0, &mut used);
+        result
+    }
+}
+```
+
 ## C 
 
 ```c

problems/0056.合并区间.md

@@ -297,7 +297,37 @@ function merge(intervals: number[][]): number[][] {
 };
 ```
 
+### Scala
 
+```scala
+object Solution {
+  import scala.collection.mutable
+  def merge(intervals: Array[Array[Int]]): Array[Array[Int]] = {
+    var res = mutable.ArrayBuffer[Array[Int]]()
+
+    // 排序
+    var interval = intervals.sortWith((a, b) => {
+      a(0) < b(0)
+    })
+
+    var left = interval(0)(0)
+    var right = interval(0)(1)
+
+    for (i <- 1 until interval.length) {
+      if (interval(i)(0) <= right) {
+        left = math.min(left, interval(i)(0))
+        right = math.max(right, interval(i)(1))
+      } else {
+        res.append(Array[Int](left, right))
+        left = interval(i)(0)
+        right = interval(i)(1)
+      }
+    }
+    res.append(Array[Int](left, right))
+    res.toArray // 返回res的Array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/0059.螺旋矩阵II.md

@@ -598,5 +598,30 @@ object Solution {
   }
 }
 ```
+C#：
+```csharp
+public class Solution {
+    public int[][] GenerateMatrix(int n) {
+        int[][] answer = new int[n][];
+        for(int i = 0; i < n; i++)
+            answer[i] = new int[n];
+        int start = 0;
+        int end = n - 1;
+        int tmp = 1;
+        while(tmp < n * n)
+        {
+            for(int i = start; i < end; i++) answer[start][i] = tmp++;
+            for(int i = start; i < end; i++) answer[i][end] = tmp++;
+            for(int i = end; i > start; i--) answer[end][i] = tmp++;
+            for(int i = end; i > start; i--) answer[i][start] = tmp++;
+            start++;
+            end--;
+        }
+        if(n % 2 == 1) answer[n / 2][n / 2] = tmp;
+        return answer;
+    }
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0062.不同路径.md

@@ -436,5 +436,21 @@ int uniquePaths(int m, int n){
 }
 ```
 
+### Scala
+
+```scala
+object Solution {
+  def uniquePaths(m: Int, n: Int): Int = {
+    var dp = Array.ofDim[Int](m, n)
+    for (i <- 0 until m) dp(i)(0) = 1
+    for (j <- 1 until n) dp(0)(j) = 1
+    for (i <- 1 until m; j <- 1 until n) {
+      dp(i)(j) = dp(i - 1)(j) + dp(i)(j - 1)
+    }
+    dp(m - 1)(n - 1)
+  }
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0063.不同路径II.md

@@ -476,5 +476,37 @@ int uniquePathsWithObstacles(int** obstacleGrid, int obstacleGridSize, int* obst
 }
 ```
 
+### Scala
+
+```scala
+object Solution {
+  import scala.util.control.Breaks._
+  def uniquePathsWithObstacles(obstacleGrid: Array[Array[Int]]): Int = {
+    var (m, n) = (obstacleGrid.length, obstacleGrid(0).length)
+    var dp = Array.ofDim[Int](m, n)
+
+    // 比如break、continue这些流程控制需要使用breakable
+    breakable(
+      for (i <- 0 until m) {
+        if (obstacleGrid(i)(0) != 1) dp(i)(0) = 1
+        else break()
+      }
+    )
+    breakable(
+      for (j <- 0 until n) {
+        if (obstacleGrid(0)(j) != 1) dp(0)(j) = 1
+        else break()
+      }
+    )
+
+    for (i <- 1 until m; j <- 1 until n; if obstacleGrid(i)(j) != 1) {
+      dp(i)(j) = dp(i - 1)(j) + dp(i)(j - 1)
+    }
+
+    dp(m - 1)(n - 1)
+  }
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0070.爬楼梯.md

@@ -401,6 +401,38 @@ int climbStairs(int n){
 }
 ```
 
+### Scala
+
+```scala
+object Solution {
+  def climbStairs(n: Int): Int = {
+    if (n <= 2) return n
+    var dp = new Array[Int](n + 1)
+    dp(1) = 1
+    dp(2) = 2
+    for (i <- 3 to n) {
+      dp(i) = dp(i - 1) + dp(i - 2)
+    }
+    dp(n)
+  }
+}
+```
+
+优化空间复杂度:
+```scala
+object Solution {
+  def climbStairs(n: Int): Int = {
+    if (n <= 2) return n
+    var (a, b) = (1, 2)
+    for (i <- 3 to n) {
+      var tmp = a + b
+      a = b
+      b = tmp
+    }
+    b // 最终返回b
+  }
+}
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>