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

problems/0001.两数之和.md

@@ -207,18 +207,16 @@ function twoSum(array $nums, int $target): array
 Swift：
 ```swift
 func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
-    var res = [Int]()
+    // 值: 下标
-    var dict = [Int : Int]()
+    var map = [Int: Int]()
-    for i in 0 ..< nums.count {
+    for (i, e) in nums.enumerated() {
-        let other = target - nums[i]
+        if let v = map[target - e] {
-        if dict.keys.contains(other) {
+            return [v, i]
-            res.append(i)
+        } else {
-            res.append(dict[other]!)
+            map[e] = i
-            return res
         }
-        dict[nums[i]] = i
     }
-    return res
+    return []
 }
 ```
 

problems/0027.移除元素.md

@@ -197,7 +197,23 @@ var removeElement = (nums, val) => {
 };
 ```
 
+TypeScript：
+
+```typescript
+function removeElement(nums: number[], val: number): number {
+    let slowIndex: number = 0, fastIndex: number = 0;
+    while (fastIndex < nums.length) {
+        if (nums[fastIndex] !== val) {
+            nums[slowIndex++] = nums[fastIndex];
+        }
+        fastIndex++;
+    }
+    return slowIndex;
+};
+```
+
 Ruby:
+
 ```ruby
 def remove_element(nums, val)
     i = 0

problems/0028.实现strStr.md

@@ -649,6 +649,41 @@ class Solution {
 }
 ```
 
+```Java
+class Solution {
+    //前缀表（不减一）Java实现
+    public int strStr(String haystack, String needle) {
+        if (needle.length() == 0) return 0;
+        int[] next = new int[needle.length()];
+        getNext(next, needle);
+
+        int j = 0;
+        for (int i = 0; i < haystack.length(); i++) {
+            while (j > 0 && needle.charAt(j) != haystack.charAt(i)) 
+                j = next[j - 1];
+            if (needle.charAt(j) == haystack.charAt(i)) 
+                j++;
+            if (j == needle.length()) 
+                return i - needle.length() + 1;
+        }
+        return -1;
+
+    }
+    
+    private void getNext(int[] next, String s) {
+        int j = 0;
+        next[0] = 0;
+        for (int i = 1; i < s.length(); i++) {
+            while (j > 0 && s.charAt(j) != s.charAt(i)) 
+                j = next[j - 1];
+            if (s.charAt(j) == s.charAt(i)) 
+                j++;
+            next[i] = j; 
+        }
+    }
+}
+```
+
 Python3：
 
 ```python

problems/0031.下一个排列.md

@@ -86,7 +86,7 @@ public:
                 }
             }
         }
-        // 到这里了说明整个数组都是倒叙了，反转一下便可
+        // 到这里了说明整个数组都是倒序了，反转一下便可
         reverse(nums.begin(), nums.end());
     }
 };

problems/0040.组合总和II.md

@@ -255,6 +255,7 @@ public:
 
 
 ## Java 
+**使用标记数组**
 ```Java
 class Solution {
     List<List<Integer>> lists = new ArrayList<>();
@@ -292,6 +293,44 @@ class Solution {
     }
 }
 ```
+**不使用标记数组**
+```Java
+class Solution {
+  List<List<Integer>> res = new ArrayList<>();
+  LinkedList<Integer> path = new LinkedList<>();
+  int sum = 0;
+  
+  public List<List<Integer>> combinationSum2( int[] candidates, int target ) {
+    //为了将重复的数字都放到一起，所以先进行排序
+    Arrays.sort( candidates );
+    backTracking( candidates, target, 0 );
+    return res;
+  }
+  
+  private void backTracking( int[] candidates, int target, int start ) {
+    if ( sum == target ) {
+      res.add( new ArrayList<>( path ) );
+      return;
+    }
+    for ( int i = start; i < candidates.length && sum + candidates[i] <= target; i++ ) {
+      //正确剔除重复解的办法
+      //跳过同一树层使用过的元素
+      if ( i > start && candidates[i] == candidates[i - 1] ) {
+        continue;
+      }
+
+      sum += candidates[i];
+      path.add( candidates[i] );
+      // i+1 代表当前组内元素只选取一次
+      backTracking( candidates, target, i + 1 );
+
+      int temp = path.getLast();
+      sum -= temp;
+      path.removeLast();
+    }
+  }
+}
+```
 
 ## Python 
 **回溯+巧妙去重(省去使用used**
@@ -384,6 +423,7 @@ class Solution:
 ## Go
 主要在于如何在回溯中去重
 
+**使用used数组**
 ```go
 func combinationSum2(candidates []int, target int) [][]int {
     var trcak []int
@@ -423,7 +463,41 @@ func backtracking(startIndex,sum,target int,candidates,trcak []int,res *[][]int,
     }
 }
 ```
-
+**不使用used数组**
+```go
+func combinationSum2(candidates []int, target int) [][]int {
+    var trcak []int
+    var res [][]int
+    sort.Ints(candidates)
+    backtracking(0,0,target,candidates,trcak,&res)
+    return res
+}
+func backtracking(startIndex,sum,target int,candidates,trcak []int,res *[][]int){
+    //终止条件
+    if sum==target{
+        tmp:=make([]int,len(trcak))
+        //拷贝
+        copy(tmp,trcak)
+        //放入结果集
+        *res=append(*res,tmp)
+        return
+    }
+    //回溯
+    for i:=startIndex;i<len(candidates) && sum+candidates[i]<=target;i++{
+        // 若当前树层有使用过相同的元素，则跳过
+        if i>startIndex&&candidates[i]==candidates[i-1]{
+                continue
+        }
+        //更新路径集合和sum
+        trcak=append(trcak,candidates[i])
+        sum+=candidates[i]
+        backtracking(i+1,sum,target,candidates,trcak,res)
+        //回溯
+        trcak=trcak[:len(trcak)-1]
+        sum-=candidates[i]
+    }
+}
+```
 ## javaScript
 
 ```js

problems/0047.全排列II.md

@@ -323,5 +323,76 @@ func permuteUnique(_ nums: [Int]) -> [[Int]] {
 }
 ```
 
+### C
+```c
+//临时数组
+int *path;
+//返回数组
+int **ans;
+int *used;
+int pathTop, ansTop;
+
+//拷贝path到ans中
+void copyPath() {
+    int *tempPath = (int*)malloc(sizeof(int) * pathTop);
+    int i;
+    for(i = 0; i < pathTop; ++i) {
+        tempPath[i] = path[i];
+    }
+    ans[ansTop++] = tempPath;
+}
+
+void backTracking(int* used, int *nums, int numsSize) {
+    //若path中元素个数等于numsSize，将path拷贝入ans数组中
+    if(pathTop == numsSize)
+        copyPath();
+    int i;
+    for(i = 0; i < numsSize; i++) {
+        //若当前元素已被使用
+        //或前一位元素与当前元素值相同但并未被使用
+        //则跳过此分支
+        if(used[i] || (i != 0 && nums[i] == nums[i-1] && used[i-1] == 0))
+            continue;
+
+        //将当前元素的使用情况设为True
+        used[i] = 1;
+        path[pathTop++] = nums[i];
+        backTracking(used, nums, numsSize);
+        used[i] = 0;
+        --pathTop;
+    }
+}
+
+int cmp(void* elem1, void* elem2) {
+    return *((int*)elem1) - *((int*)elem2);
+}
+
+int** permuteUnique(int* nums, int numsSize, int* returnSize, int** returnColumnSizes){
+    //排序数组
+    qsort(nums, numsSize, sizeof(int), cmp);
+    //初始化辅助变量
+    pathTop = ansTop = 0;
+    path = (int*)malloc(sizeof(int) * numsSize);
+    ans = (int**)malloc(sizeof(int*) * 1000);
+    //初始化used辅助数组
+    used = (int*)malloc(sizeof(int) * numsSize);
+    int i;
+    for(i = 0; i < numsSize; i++) {
+        used[i] = 0;
+    }
+
+    backTracking(used, nums, numsSize);
+
+    //设置返回的数组的长度
+    *returnSize = ansTop;
+    *returnColumnSizes = (int*)malloc(sizeof(int) * ansTop);
+    int z;
+    for(z = 0; z < ansTop; z++) {
+        (*returnColumnSizes)[z] = numsSize;
+    }
+    return ans;
+}
+```
+
 -----------------------
 <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

@@ -166,7 +166,7 @@ if (i > startIndex && nums[i] == nums[i - 1] ) {
 
 
 ### Java  
-
+使用used数组
 ```java
 class Solution {
    List<List<Integer>> result = new ArrayList<>();// 存放符合条件结果的集合
@@ -202,6 +202,37 @@ class Solution {
 }
 ```
 
+不使用used数组
+```java
+class Solution {
+
+  List<List<Integer>> res = new ArrayList<>();
+  LinkedList<Integer> path = new LinkedList<>();
+  
+  public List<List<Integer>> subsetsWithDup( int[] nums ) {
+    Arrays.sort( nums );
+    subsetsWithDupHelper( nums, 0 );
+    return res;
+  }
+
+
+  private void subsetsWithDupHelper( int[] nums, int start ) {
+    res.add( new ArrayList<>( path ) );
+
+    for ( int i = start; i < nums.length; i++ ) {
+        // 跳过当前树层使用过的、相同的元素
+      if ( i > start && nums[i - 1] == nums[i] ) {
+        continue;
+      }
+      path.add( nums[i] );
+      subsetsWithDupHelper( nums, i + 1 );
+      path.removeLast();
+    }
+  }
+
+}
+```
+
 ### Python 
 ```python
 class Solution:

problems/0101.对称二叉树.md

@@ -574,7 +574,87 @@ var isSymmetric = function(root) {
 };
 ```
 
+## Swift:
 
+> 递归
+```swift
+func isSymmetric(_ root: TreeNode?) -> Bool {
+    return _isSymmetric(root?.left, right: root?.right)
+}
+func _isSymmetric(_ left: TreeNode?, right: TreeNode?) -> Bool {
+    // 首先排除空节点情况
+    if left == nil && right == nil {
+        return true
+    } else if left == nil && right != nil {
+        return false
+    } else if left != nil && right == nil {
+        return false
+    } else if left!.val != right!.val {
+        // 进而排除数值不相等的情况
+        return false
+    }
+    
+    // left 和 right 都不为空, 且数值也相等就递归
+    let inSide = _isSymmetric(left!.right, right: right!.left)
+    let outSide = _isSymmetric(left!.left, right: right!.right)
+    return inSide && outSide
+}
+```
+
+> 迭代 - 使用队列
+```swift
+func isSymmetric2(_ root: TreeNode?) -> Bool {
+    guard let root = root else {
+        return true
+    }
+    var queue = [TreeNode?]()
+    queue.append(root.left)
+    queue.append(root.right)
+    while !queue.isEmpty {
+        let left = queue.removeFirst()
+        let right = queue.removeFirst()
+        if left == nil && right == nil {
+            continue
+        }
+        if left == nil || right == nil || left?.val != right?.val {
+            return false
+        }
+        queue.append(left!.left)
+        queue.append(right!.right)
+        queue.append(left!.right)
+        queue.append(right!.left)
+    }
+    return true
+}
+```
+
+> 迭代 - 使用栈
+```swift
+func isSymmetric3(_ root: TreeNode?) -> Bool {
+    guard let root = root else {
+        return true
+    }
+    var stack = [TreeNode?]()
+    stack.append(root.left)
+    stack.append(root.right)
+    while !stack.isEmpty {
+        let left = stack.removeLast()
+        let right = stack.removeLast()
+        
+        if left == nil && right == nil {
+            continue
+        }
+        if left == nil || right == nil || left?.val != right?.val {
+            return false
+        }
+        stack.append(left!.left)
+        stack.append(right!.right)
+        stack.append(left!.right)
+        stack.append(right!.left)
+    }
+    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/0102.二叉树的层序遍历.md

@@ -246,6 +246,34 @@ var levelOrder = function(root) {
 
 ```
 
+Swift:
+```swift
+func levelOrder(_ root: TreeNode?) -> [[Int]] {
+    var res = [[Int]]()
+    guard let root = root else {
+        return res
+    }
+    var queue = [TreeNode]()
+    queue.append(root)
+    while !queue.isEmpty {
+        let size = queue.count
+        var sub = [Int]()
+        for _ in 0 ..< size {
+            let node = queue.removeFirst()
+            sub.append(node.val)
+            if let left = node.left {
+                queue.append(left)
+            }
+            if let right = node.right {
+                queue.append(right)
+            }
+        }
+        res.append(sub)
+    }
+    return res
+}
+```
+
 **此时我们就掌握了二叉树的层序遍历了，那么如下九道力扣上的题目，只需要修改模板的两三行代码（不能再多了），便可打倒！**
 
 
@@ -426,6 +454,31 @@ var levelOrderBottom = function(root) {
 };
 ```
 
+Swift:
+```swift
+func levelOrderBottom(_ root: TreeNode?) -> [[Int]] {
+    var res =  [[Int]]()
+    guard let root = root else {
+        return res
+    }
+    var queue:  [TreeNode] = [root]
+    while !queue.isEmpty {
+        var sub = [Int]()
+        for _ in 0 ..< queue.count {
+            let node = queue.removeFirst()
+            sub.append(node.val)
+            if let left = node.left {
+                queue.append(left)
+            }
+            if let right = node.right {
+                queue.append(right)
+            }
+        }
+        res.insert(sub, at: 0)
+    }
+    return res
+}
+```
 
 # 199.二叉树的右视图
 
@@ -604,6 +657,36 @@ var rightSideView = function(root) {
 };
 ```
 
+Swift:
+```swift
+func rightSideView(_ root: TreeNode?) -> [Int] {
+    var res = [Int]()
+    guard let root = root else {
+        return res
+    }
+    var queue = [TreeNode]()
+    queue.append(root)
+    while !queue.isEmpty {
+        let size = queue.count
+        for i in 0 ..< size {
+            let node = queue.removeFirst()
+            if i == size - 1 {
+                // 保存 每层最后一个元素
+                res.append(node.val)
+            }
+            if let left = node.left {
+                queue.append(left)
+            }
+            if let right = node.right {
+                queue.append(right)
+            }
+        }
+    }
+    return res
+}
+```
+
+
 # 637.二叉树的层平均值
 
 [力扣题目链接](https://leetcode-cn.com/problems/average-of-levels-in-binary-tree/)
@@ -785,6 +868,34 @@ var averageOfLevels = function(root) {
 };
 ```
 
+Swift:
+```swift
+func averageOfLevels(_ root: TreeNode?) -> [Double] {
+    var res = [Double]()
+    guard let root = root else {
+        return res
+    }
+    var queue = [TreeNode]()
+    queue.append(root)
+    while !queue.isEmpty {
+        let size = queue.count
+        var sum = 0
+        for _ in 0 ..< size {
+            let node = queue.removeFirst()
+            sum += node.val
+            if let left = node.left {
+                queue.append(left)
+            }
+            if let right = node.right {
+                queue.append(right)
+            }
+        }
+        res.append(Double(sum) / Double(size))
+    }
+    return res
+}
+```
+
 # 429.N叉树的层序遍历
 
 [力扣题目链接](https://leetcode-cn.com/problems/n-ary-tree-level-order-traversal/)
@@ -981,6 +1092,31 @@ var levelOrder = function(root) {
 };
 ```
 
+Swift:
+```swift
+func levelOrder(_ root: Node?) -> [[Int]] {
+    var res = [[Int]]()
+    guard let root = root else {
+        return res
+    }
+    var queue = [Node]()
+    queue.append(root)
+    while !queue.isEmpty {
+        let size = queue.count
+        var sub = [Int]()
+        for _ in 0 ..< size {
+            let node = queue.removeFirst()
+            sub.append(node.val)
+            for childNode in node.children {
+                queue.append(childNode)
+            }
+        }
+        res.append(sub)
+    }
+    return res
+}
+```
+
 # 515.在每个树行中找最大值
 
 [力扣题目链接](https://leetcode-cn.com/problems/find-largest-value-in-each-tree-row/)
@@ -1136,6 +1272,36 @@ var largestValues = function(root) {
 };
 ```
 
+Swift:
+```swift
+func largestValues(_ root: TreeNode?) -> [Int] {
+    var res = [Int]()
+    guard let root = root else {
+        return res
+    }
+    var queue = [TreeNode]()
+    queue.append(root)
+    while !queue.isEmpty {
+        let size = queue.count
+        var max: Int = Int.min
+        for _ in 0 ..< size {
+            let node = queue.removeFirst()
+            if node.val > max {
+                max = node.val
+            }
+            if let left = node.left {
+                queue.append(left)
+            }
+            if let right = node.right {
+                queue.append(right)
+            }
+        }
+        res.append(max)
+    }
+    return res
+}
+```
+
 # 116.填充每个节点的下一个右侧节点指针
 
 [力扣题目链接](https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node/)
@@ -1338,6 +1504,37 @@ func connect(root *Node) *Node {
 }
 ```
 
+Swift:
+```swift
+func connect(_ root: Node?) -> Node? {
+    guard let root = root else {
+        return nil
+    }
+    var queue = [Node]()
+    queue.append(root)
+    while !queue.isEmpty {
+        let size = queue.count
+        var preNode: Node?
+        for i in 0 ..< size {
+            let node = queue.removeFirst()
+            if i == 0 {
+                preNode = node
+            } else {
+                preNode?.next = node
+                preNode = node
+            }
+            if let left = node.left {
+                queue.append(left)
+            }
+            if let right = node.right {
+                queue.append(right)
+            }
+        }
+    }
+    return root
+}
+```
+
 # 117.填充每个节点的下一个右侧节点指针II
 
 [力扣题目链接](https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node-ii/)
@@ -1532,6 +1729,38 @@ func connect(root *Node) *Node {
     return root
 }
 ```
+
+Swift：
+```swift
+func connect(_ root: Node?) -> Node? {
+    guard let root = root else {
+        return nil
+    }
+    var queue = [Node]()
+    queue.append(root)
+    while !queue.isEmpty {
+        let size = queue.count
+        var preNode: Node?
+        for i in 0 ..< size {
+            let node = queue.removeFirst()
+            if i == 0 {
+                preNode = node
+            } else {
+                preNode?.next = node
+                preNode = node
+            }
+            if let left = node.left {
+                queue.append(left)
+            }
+            if let right = node.right {
+                queue.append(right)
+            }
+        }
+    }
+    return root
+}
+```
+
 # 104.二叉树的最大深度
 
 [力扣题目链接](https://leetcode-cn.com/problems/maximum-depth-of-binary-tree/)
@@ -1704,6 +1933,31 @@ var maxDepth = function(root) {
 };
 ```
 
+Swift:
+```swift
+func maxDepth(_ root: TreeNode?) -> Int {
+    guard let root = root else {
+        return 0
+    }
+    var queue = [TreeNode]()
+    queue.append(root)
+    var res: Int = 0
+    while !queue.isEmpty {
+        for _ in 0 ..< queue.count {
+            let node = queue.removeFirst()
+            if let left = node.left {
+                queue.append(left)
+            }
+            if let right = node.right {
+                queue.append(right)
+            }
+        }
+        res += 1
+    }
+    return res
+}
+```
+
 # 111.二叉树的最小深度
 
 [力扣题目链接](https://leetcode-cn.com/problems/minimum-depth-of-binary-tree/)
@@ -1876,7 +2130,33 @@ var minDepth = function(root) {
 };
 ```
 
-
+Swift：
+```swift
+func minDepth(_ root: TreeNode?) -> Int {
+    guard let root = root else {
+        return 0
+    }
+    var res = 0
+    var queue = [TreeNode]()
+    queue.append(root)
+    while !queue.isEmpty {
+        res += 1
+        for _ in 0 ..< queue.count {
+            let node = queue.removeFirst()
+            if node.left == nil && node.right == nil {
+                return res
+            }
+            if let left = node.left {
+                queue.append(left)
+            }
+            if let right = node.right {
+                queue.append(right)
+            }
+        }
+    }
+    return res
+}
+```
 
 
 # 总结

problems/0104.二叉树的最大深度.md

@@ -653,5 +653,82 @@ int maxDepth(struct TreeNode* root){
 }
 ```
 
+## Swift
+
+>二叉树最大深度
+```swift
+// 递归 - 后序
+func maxDepth1(_ root: TreeNode?) -> Int {
+    return _maxDepth1(root)
+}
+func _maxDepth1(_ root: TreeNode?) -> Int {
+    if root == nil {
+        return 0
+    }
+    let leftDepth = _maxDepth1(root!.left)
+    let rightDepth = _maxDepth1(root!.right)
+    return 1 + max(leftDepth, rightDepth)
+}
+
+// 层序
+func maxDepth(_ root: TreeNode?) -> Int {
+    guard let root = root else {
+        return 0
+    }
+    var queue = [TreeNode]()
+    queue.append(root)
+    var res: Int = 0
+    while !queue.isEmpty {
+        res += 1
+        for _ in 0 ..< queue.count {
+            let node = queue.removeFirst()
+            if let left = node.left {
+                queue.append(left)
+            }
+            if let right = node.right {
+                queue.append(right)
+            }
+        }
+    }
+    return res
+}
+```
+
+>N叉树最大深度
+```swift
+// 递归
+func maxDepth(_ root: Node?) -> Int {
+    guard let root = root else {
+        return 0
+    }
+    var depth = 0
+    for node in root.children {
+        depth = max(depth, maxDepth(node))
+    }
+    return depth + 1
+}
+
+// 迭代-层序遍历
+func maxDepth1(_ root: Node?) -> Int {
+    guard let root = root else {
+        return 0
+    }
+    var depth = 0
+    var queue = [Node]()
+    queue.append(root)
+    while !queue.isEmpty {
+        let size = queue.count
+        depth += 1
+        for _ in 0 ..< size {
+            let node = queue.removeFirst()
+            for child in node.children {
+                queue.append(child)
+            }
+        }
+    }
+    return depth
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0106.从中序与后序遍历序列构造二叉树.md

@@ -816,6 +816,7 @@ var buildTree = function(preorder, inorder) {
 
 ## C
 106 从中序与后序遍历序列构造二叉树
+
 ```c
 int linearSearch(int* arr, int arrSize, int key) {
     int i;
@@ -847,6 +848,7 @@ struct TreeNode* buildTree(int* inorder, int inorderSize, int* postorder, int po
 ```
 
 105 从前序与中序遍历序列构造二叉树
+
 ```c
 struct TreeNode* buildTree(int* preorder, int preorderSize, int* inorder, int inorderSize){
     // 递归结束条件：传入的数组大小为0
@@ -889,5 +891,100 @@ struct TreeNode* buildTree(int* preorder, int preorderSize, int* inorder, int in
 }
 ```
 
+## Swift
+
+105 从前序与中序遍历序列构造二叉树
+
+```swift
+class Solution {
+    func buildTree(_ preorder: [Int], _ inorder: [Int]) -> TreeNode? {
+        return helper(preorder: preorder,
+                      preorderBegin: 0,
+                      preorderEnd: preorder.count,
+                      inorder: inorder,
+                      inorderBegin: 0,
+                      inorderEnd: inorder.count)
+    }
+    
+    func helper(preorder: [Int], preorderBegin: Int, preorderEnd: Int, inorder: [Int], inorderBegin: Int, inorderEnd: Int) -> TreeNode? {
+        if preorderBegin == preorderEnd {
+            return nil
+        }
+        
+        // 前序遍历数组的第一个元素作为分割点
+        let rootValue = preorder[preorderBegin]
+        let root = TreeNode(rootValue)
+        
+        
+        if preorderEnd - preorderBegin == 1 {
+            return root
+        }
+        
+        var index = 0 // 从中序遍历数组中找到根节点的下标
+        if let ind = inorder.firstIndex(of: rootValue) {
+            index = ind
+        }
+        
+        // 递归
+        root.left = helper(preorder: preorder,
+                           preorderBegin: preorderBegin + 1,
+                           preorderEnd: preorderBegin + 1 + index - inorderBegin,
+                           inorder: inorder,
+                           inorderBegin: inorderBegin,
+                           inorderEnd: index)
+        root.right = helper(preorder: preorder,
+                            preorderBegin: preorderBegin + 1 + index - inorderBegin,
+                            preorderEnd: preorderEnd,
+                            inorder: inorder,
+                            inorderBegin: index + 1,
+                            inorderEnd: inorderEnd)
+        return root
+    }
+}
+```
+
+106 从中序与后序遍历序列构造二叉树
+
+```swift
+class Solution_0106 {
+    func buildTree(inorder: [Int], inorderBegin: Int, inorderEnd: Int, postorder: [Int], postorderBegin: Int, postorderEnd: Int) -> TreeNode? {
+        if postorderEnd - postorderBegin < 1 {
+            return nil
+        }
+        
+        // 后序遍历数组的最后一个元素作为分割点
+        let rootValue = postorder[postorderEnd - 1]
+        let root = TreeNode(rootValue)
+        
+        if postorderEnd - postorderBegin == 1 {
+            return root
+        }
+        
+        // 从中序遍历数组中找到根节点的下标
+        var delimiterIndex = 0
+        if let index = inorder.firstIndex(of: rootValue) {
+            delimiterIndex = index
+        }
+        
+        root.left = buildTree(inorder: inorder,
+                              inorderBegin: inorderBegin,
+                              inorderEnd: delimiterIndex,
+                              postorder: postorder,
+                              postorderBegin: postorderBegin,
+                              postorderEnd: postorderBegin + (delimiterIndex - inorderBegin))
+        
+        root.right = buildTree(inorder: inorder,
+                               inorderBegin: delimiterIndex + 1,
+                               inorderEnd: inorderEnd,
+                               postorder: postorder,
+                               postorderBegin: postorderBegin + (delimiterIndex - inorderBegin),
+                               postorderEnd: postorderEnd - 1)
+        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/0111.二叉树的最小深度.md

@@ -404,8 +404,51 @@ var minDepth = function(root) {
 };
 ```
 
+## Swift
 
+> 递归
+```Swift
+func minDepth(_ root: TreeNode?) -> Int {
+    guard let root = root else {
+        return 0
+    }
+    if root.left == nil && root.right != nil {
+        return 1 + minDepth(root.right)
+    }
+    if root.left != nil && root.right == nil {
+        return 1 + minDepth(root.left)
+    }
+    return 1 + min(minDepth(root.left), minDepth(root.right))
+}
+```
 
+> 迭代
+```Swift
+func minDepth(_ root: TreeNode?) -> Int {
+    guard let root = root else {
+        return 0
+    }
+    var res = 0
+    var queue = [TreeNode]()
+    queue.append(root)
+    while !queue.isEmpty {
+        res += 1
+        for _ in 0 ..< queue.count {
+            let node = queue.removeFirst()
+            if node.left == nil && node.right == nil {
+                return res
+            }
+            if let left = node.left {
+                queue.append(left)
+            }
+            if let right = node.right {
+                queue.append(right)
+            }
+        }
+    }
+    return res
+}
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0131.分割回文串.md

@@ -587,5 +587,57 @@ func partition(_ s: String) -> [[String]] {
 }
 ```
 
+## Rust
+
+```rust
+impl Solution {
+    pub fn partition(s: String) -> Vec<Vec<String>> {
+        let mut ret = vec![];
+        let mut path = vec![];
+        let sub_str: Vec<char> = s.chars().collect();
+        
+        Self::backtracing(&sub_str, 0, &mut ret, &mut path);
+        
+        ret
+    }
+    
+    fn backtracing(sub_str: &Vec<char>, start: usize, ret: &mut Vec<Vec<String>>, path: &mut Vec<String>) {
+        //如果起始位置大于s的大小，说明找到了一组分割方案
+        if start >= sub_str.len() {
+            ret.push(path.clone());
+            return;
+        }
+        
+        for i in start..sub_str.len() {
+            if !Self::is_palindrome(sub_str, start, i) {
+                continue;
+            }
+            //如果是回文子串，则记录
+            let s: String = sub_str[start..i+1].into_iter().collect();
+            path.push(s);
+            
+            //起始位置后移，保证不重复
+            Self::backtracing(sub_str, i+1, ret, path); 
+            path.pop();
+        }
+        
+    }
+    
+    fn is_palindrome(s: &Vec<char>, start: usize, end: usize) -> bool {
+        let (mut start, mut end) = (start, end);
+        
+        while start < end {
+            if s[start] != s[end] {
+                return false;
+            }
+            
+            start += 1;
+            end -= 1;
+        }
+
+        true
+    }
+}
+```
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0150.逆波兰表达式求值.md

@@ -133,39 +133,26 @@ public:
 java:
 
 ```Java
-public class EvalRPN {
+class Solution {
-
     public int evalRPN(String[] tokens) {
         Deque<Integer> stack = new LinkedList();
-        for (String token : tokens) {
+        for (int i = 0; i < tokens.length; ++i) {
-            char c = token.charAt(0);
+            if ("+".equals(tokens[i])) {        // leetcode 内置jdk的问题，不能使用==判断字符串是否相等
-            if (!isOpe(token)) {
+                stack.push(stack.pop() + stack.pop());      // 注意 - 和/ 需要特殊处理
-                stack.addFirst(stoi(token));
+            } else if ("-".equals(tokens[i])) {
-            } else if (c == '+') {
-                stack.push(stack.pop() + stack.pop());
-            } else if (c == '-') {
                 stack.push(-stack.pop() + stack.pop());
-            } else if (c == '*') {
+            } else if ("*".equals(tokens[i])) {
                 stack.push(stack.pop() * stack.pop());
+            } else if ("/".equals(tokens[i])) {
+                int temp1 = stack.pop();
+                int temp2 = stack.pop();
+                stack.push(temp2 / temp1);
             } else {
-                int num1 = stack.pop();
+                stack.push(Integer.valueOf(tokens[i]));
-                int num2 = stack.pop();
-                stack.push( num2/num1);
             }
         }
         return stack.pop();
     }
-    private boolean isOpe(String s) {
-        return s.length() == 1 && s.charAt(0) <'0' || s.charAt(0) >'9';
-    }
-    private int stoi(String s) {
-        return Integer.valueOf(s);
-    }
-
-    public static void main(String[] args) {
-        new EvalRPN().evalRPN(new String[] {"10","6","9","3","+","-11","*","/","*","17","+","5","+"});
-    }
-
 }
 ```
 

problems/0151.翻转字符串里的单词.md

@@ -40,7 +40,7 @@
 
 不能使用辅助空间之后，那么只能在原字符串上下功夫了。
 
-想一下，我们将整个字符串都反转过来，那么单词的顺序指定是倒序了，只不过单词本身也倒叙了，那么再把单词反转一下，单词不就正过来了。
+想一下，我们将整个字符串都反转过来，那么单词的顺序指定是倒序了，只不过单词本身也倒序了，那么再把单词反转一下，单词不就正过来了。
 
 所以解题思路如下：
 

problems/0203.移除链表元素.md

@@ -302,7 +302,63 @@ var removeElements = function(head, val) {
 };
 ```
 
+TypeScript:
+
+版本一（在原链表上直接删除）：
+
+```typescript
+/**
+ * Definition for singly-linked list.
+ * class ListNode {
+ *     val: number
+ *     next: ListNode | null
+ *     constructor(val?: number, next?: ListNode | null) {
+ *         this.val = (val===undefined ? 0 : val)
+ *         this.next = (next===undefined ? null : next)
+ *     }
+ * }
+ */
+function removeElements(head: ListNode | null, val: number): ListNode | null {
+    // 删除头部节点
+    while (head !== null && head.val === val) {
+        head = head.next;
+    }
+    if (head === null) return head;
+    let pre: ListNode = head, cur: ListNode = head.next;
+    // 删除非头部节点
+    while (cur) {
+        if (cur.val === val) {
+            pre.next = cur.next;
+        } else {
+            pre = pre.next;
+        }
+        cur = cur.next;
+    }
+    return head;
+};
+```
+
+版本二（虚拟头节点）：
+
+```typescript
+function removeElements(head: ListNode | null, val: number): ListNode | null {
+    head = new ListNode(0, head);
+    let pre: ListNode = head, cur: ListNode = head.next;
+    // 删除非头部节点
+    while (cur) {
+        if (cur.val === val) {
+            pre.next = cur.next;
+        } else {
+            pre = pre.next;
+        }
+        cur = cur.next;
+    }
+    return head.next;
+};
+```
+
 Swift：
+
 ```swift
 /**
  * Definition for singly-linked list.

problems/0206.翻转链表.md

@@ -314,6 +314,54 @@ var reverseList = function(head) {
 };
 ```
 
+TypeScript:
+
+```typescript
+// 双指针法
+function reverseList(head: ListNode | null): ListNode | null {
+    let preNode: ListNode | null = null,
+        curNode: ListNode | null = head,
+        tempNode: ListNode | null;
+    while (curNode) {
+        tempNode = curNode.next;
+        curNode.next = preNode;
+        preNode = curNode;
+        curNode = tempNode;
+    }
+    return preNode;
+};
+
+// 递归（从前往后翻转）
+function reverseList(head: ListNode | null): ListNode | null {
+    function recur(preNode: ListNode | null, curNode: ListNode | null): ListNode | null {
+        if (curNode === null) return preNode;
+        let tempNode: ListNode | null = curNode.next;
+        curNode.next = preNode;
+        preNode = curNode;
+        curNode = tempNode;
+        return recur(preNode, curNode);
+    }
+    return recur(null, head);
+};
+
+// 递归（从后往前翻转）
+function reverseList(head: ListNode | null): ListNode | null {
+    if (head === null) return null;
+    let newHead: ListNode | null;
+    function recur(node: ListNode | null, preNode: ListNode | null): void {
+        if (node.next === null) {
+            newHead = node;
+            newHead.next = preNode;
+        } else {
+            recur(node.next, node);
+            node.next = preNode;
+        }
+    }
+    recur(head, null);
+    return newHead;
+};
+```
+
 Ruby:
 
 ```ruby

problems/0209.长度最小的子数组.md

@@ -121,7 +121,6 @@ public:
 
 
 
-
 ## 其他语言版本
 
 
@@ -214,6 +213,28 @@ var minSubArrayLen = function(target, nums) {
 };
 ```
 
+Typescript：
+
+```typescript
+function minSubArrayLen(target: number, nums: number[]): number {
+    let left: number = 0, right: number = 0;
+    let res: number = nums.length + 1;
+    let sum: number = 0;
+    while (right < nums.length) {
+        sum += nums[right];
+        if (sum >= target) {
+            // 不断移动左指针，直到不能再缩小为止
+            while (sum - nums[left] >= target) {
+                sum -= nums[left++];
+            }
+            res = Math.min(res, right - left + 1);
+        }
+        right++;
+    }
+    return res === nums.length + 1 ? 0 : res;
+};
+```
+
 Swift:
 
 ```swift
@@ -291,5 +312,23 @@ class Solution {
 }
 ```
 
+Ruby:
+
+```ruby
+def min_sub_array_len(target, nums)
+  res = Float::INFINITY # 无穷大
+  i, sum = 0, 0
+  nums.length.times do |j|
+    sum	+= nums[j]
+    while sum >= target
+      res = [res, j - i + 1].min
+      sum -= nums[i]
+      i	+= 1
+    end
+  end
+  res == Float::INFINITY ? 0 : res
+end
+```
+
 -----------------------
 <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

@@ -323,7 +323,6 @@ class Solution:
             self.backtracking(k, n, i + 1)
             self.path.pop()
             self.sum_now -= i
-        return
 ```
 
 ## Go

problems/0222.完全二叉树的节点个数.md

@@ -80,7 +80,7 @@ return treeNum;
 class Solution {
 private:
     int getNodesNum(TreeNode* cur) {
-        if (cur == 0) return 0;
+        if (cur == NULL) return 0;
         int leftNum = getNodesNum(cur->left);      // 左
         int rightNum = getNodesNum(cur->right);    // 右
         int treeNum = leftNum + rightNum + 1;      // 中
@@ -522,5 +522,73 @@ int countNodes(struct TreeNode* root){
 }
 ```
 
+## Swift:
+
+> 递归
+```swift
+func countNodes(_ root: TreeNode?) -> Int {
+    return _countNodes(root)
+}
+func _countNodes(_ root: TreeNode?) -> Int {
+    guard let root = root else {
+        return 0
+    }
+    let leftCount = _countNodes(root.left)
+    let rightCount = _countNodes(root.right)
+    return 1 + leftCount + rightCount
+}
+```
+  
+> 层序遍历
+```Swift
+func countNodes(_ root: TreeNode?) -> Int {
+    guard let root = root else {
+        return 0
+    }
+    var res = 0
+    var queue = [TreeNode]()
+    queue.append(root)
+    while !queue.isEmpty {
+        let size = queue.count
+        for _ in 0 ..< size {
+            let node = queue.removeFirst()
+            res += 1
+            if let left = node.left {
+                queue.append(left)
+            }
+            if let right = node.right {
+                queue.append(right)
+            }
+        }
+    }
+    return res
+}
+```
+  
+> 利用完全二叉树性质
+```Swift
+func countNodes(_ root: TreeNode?) -> Int {
+    guard let root = root else {
+        return 0
+    }
+    var leftNode = root.left
+    var rightNode = root.right
+    var leftDepth = 0
+    var rightDepth = 0
+    while leftNode != nil {
+        leftNode = leftNode!.left
+        leftDepth += 1
+    }
+    while rightNode != nil {
+        rightNode = rightNode!.right
+        rightDepth += 1
+    }
+    if leftDepth == rightDepth {
+        return (2 << leftDepth) - 1
+    }
+    return countNodes(root.left) + countNodes(root.right) + 1
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0226.翻转二叉树.md

@@ -609,5 +609,43 @@ struct TreeNode* invertTree(struct TreeNode* root){
 }
 ```
 
+### Swift：
+```swift
+// 前序遍历-递归
+func invertTree(_ root: TreeNode?) -> TreeNode? {
+    guard let root = root else {
+        return root
+    }
+    let tmp = root.left
+    root.left = root.right
+    root.right = tmp
+    let _ = invertTree(root.left)
+    let _ = invertTree(root.right)
+    return root
+}
+
+// 层序遍历-迭代
+func invertTree1(_ root: TreeNode?) -> TreeNode? {
+    guard let root = root else {
+        return nil
+    }
+    var queue = [TreeNode]()
+    queue.append(root)
+    while !queue.isEmpty {
+        let node = queue.removeFirst()
+        let tmp = node.left
+        node.left = node.right
+        node.right = tmp
+        if let left = node.left {
+            queue.append(left)
+        }
+        if let right = node.right {
+            queue.append(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/0309.最佳买卖股票时机含冷冻期.md

@@ -93,7 +93,7 @@ dp[i][3] = dp[i - 1][2];
 综上分析，递推代码如下：
 
 ```CPP
-dp[i][0] = max(dp[i - 1][0], max(dp[i - 1][3], dp[i - 1][1]) - prices[i];
+dp[i][0] = max(dp[i - 1][0], max(dp[i - 1][3], dp[i - 1][1]) - prices[i]);
 dp[i][1] = max(dp[i - 1][1], dp[i - 1][3]);
 dp[i][2] = dp[i - 1][0] + prices[i];
 dp[i][3] = dp[i - 1][2];

problems/0344.反转字符串.md

@@ -101,7 +101,6 @@ s[j] = tmp;
 s[i] ^= s[j];
 s[j] ^= s[i];
 s[i] ^= s[j];
-
 ```
 
 这道题目还是比较简单的，但是我正好可以通过这道题目说一说在刷题的时候，使用库函数的原则。

problems/0347.前K个高频元素.md

@@ -107,7 +107,7 @@ public:
             }
         }
 
-        // 找出前K个高频元素，因为小顶堆先弹出的是最小的，所以倒叙来输出到数组
+        // 找出前K个高频元素，因为小顶堆先弹出的是最小的，所以倒序来输出到数组
         vector<int> result(k);
         for (int i = k - 1; i >= 0; i--) {
             result[i] = pri_que.top().first;
@@ -142,7 +142,7 @@ 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());
+        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) {
@@ -180,7 +180,7 @@ class Solution:
             if len(pri_que) > k: #如果堆的大小大于了K，则队列弹出，保证堆的大小一直为k
                 heapq.heappop(pri_que)
         
-        #找出前K个高频元素，因为小顶堆先弹出的是最小的，所以倒叙来输出到数组
+        #找出前K个高频元素，因为小顶堆先弹出的是最小的，所以倒序来输出到数组
         result = [0] * k
         for i in range(k-1, -1, -1):
             result[i] = heapq.heappop(pri_que)[1]

problems/0416.分割等和子集.md

@@ -112,7 +112,7 @@ vector<int> dp(10001, 0);
 
 4. 确定遍历顺序
 
-在[动态规划：关于01背包问题，你该了解这些！（滚动数组）](https://programmercarl.com/背包理论基础01背包-2.html)中就已经说明：如果使用一维dp数组，物品遍历的for循环放在外层，遍历背包的for循环放在内层，且内层for循环倒叙遍历！
+在[动态规划：关于01背包问题，你该了解这些！（滚动数组）](https://programmercarl.com/背包理论基础01背包-2.html)中就已经说明：如果使用一维dp数组，物品遍历的for循环放在外层，遍历背包的for循环放在内层，且内层for循环倒序遍历！
 
 代码如下：
 

problems/0454.四数相加II.md

@@ -229,28 +229,24 @@ class Solution {
 Swift：
 ```swift
 func fourSumCount(_ nums1: [Int], _ nums2: [Int], _ nums3: [Int], _ nums4: [Int]) -> Int {
-    // key:a+b的数值，value:a+b数值出现的次数
+    // ab和: ab和出现次数
-    var map = [Int: Int]()
+    var countDic = [Int: Int]()
-    // 遍历nums1和nums2数组，统计两个数组元素之和，和出现的次数，放到map中
+    for a in nums1 {
-    for i in 0 ..< nums1.count {
+        for b in nums2 {
-        for j in 0 ..< nums2.count {
+            let key = a + b
-            let sum1 = nums1[i] + nums2[j]
+            countDic[key] = countDic[key, default: 0] + 1
-            map[sum1] = (map[sum1] ?? 0) + 1
         }
     }
-    // 统计a+b+c+d = 0 出现的次数
+
-    var res = 0
+    // 通过-(c + d)作为key，去累加ab和出现的次数
-    // 在遍历大num3和num4数组，找到如果 0-(c+d) 在map中出现过的话，就把map中key对应的value也就是出现次数统计出来。
+    var result = 0
-    for i in 0 ..< nums3.count {
+    for c in nums3 {
-        for j in 0 ..< nums4.count {
+        for d in nums4 {
-            let sum2 = nums3[i] + nums4[j]
+            let key = -(c + d)
-            let other = 0 - sum2
+            result += countDic[key, default: 0]
-            if map.keys.contains(other) {
-                res += map[other]!
         }
     }
-    }
+    return result
-    return res
 }
 ```
 

problems/0459.重复的子字符串.md

@@ -45,7 +45,7 @@
 
 这里就要说一说next数组了，next 数组记录的就是最长相同前后缀( [字符串：KMP算法精讲](https://programmercarl.com/0028.实现strStr.html) 这里介绍了什么是前缀，什么是后缀，什么又是最长相同前后缀)， 如果 next[len - 1] != -1，则说明字符串有最长相同的前后缀（就是字符串里的前缀子串和后缀子串相同的最长长度）。
 
-最长相等前后缀的长度为：next[len - 1] + 1。
+最长相等前后缀的长度为：next[len - 1] + 1。(这里的next数组是以统一减一的方式计算的，因此需要+1)
 
 数组长度为：len。
 

problems/0674.最长连续递增序列.md

@@ -154,6 +154,8 @@ public:
 
 
 Java：
+
+> 动态规划：
 ```java
  /**
      * 1.dp[i] 代表当前下标最大连续值
@@ -180,6 +182,25 @@ Java：
     }
 ```
 
+> 贪心法：
+
+```Java
+public static int findLengthOfLCIS(int[] nums) {
+    if (nums.length == 0) return 0;
+    int res = 1; // 连续子序列最少也是1
+    int count = 1;
+    for (int i = 0; i < nums.length - 1; i++) {
+        if (nums[i + 1] > nums[i]) { // 连续记录
+            count++;
+        } else { // 不连续，count从头开始
+            count = 1;
+        }
+        if (count > res) res = count;
+    }
+    return res;
+}
+```
+
 Python：
 
 > 动态规划：

problems/0685.冗余连接II.md

@@ -68,7 +68,7 @@ for (int i = 0; i < n; i++) {
 
 ```cpp
 vector<int> vec; // 记录入度为2的边（如果有的话就两条边）
-// 找入度为2的节点所对应的边，注意要倒叙，因为优先返回最后出现在二维数组中的答案
+// 找入度为2的节点所对应的边，注意要倒序，因为优先返回最后出现在二维数组中的答案
 for (int i = n - 1; i >= 0; i--) {
     if (inDegree[edges[i][1]] == 2) {
         vec.push_back(i);
@@ -577,7 +577,7 @@ var findRedundantDirectedConnection = function(edges) {
         inDegree[edges[i][1]]++; // 统计入度
     }
     let vec = [];// 记录入度为2的边（如果有的话就两条边）
-    // 找入度为2的节点所对应的边，注意要倒叙，因为优先返回最后出现在二维数组中的答案
+    // 找入度为2的节点所对应的边，注意要倒序，因为优先返回最后出现在二维数组中的答案
     for (let i = n - 1; i >= 0; i--) {
         if (inDegree[edges[i][1]] == 2) {
             vec.push(i);

problems/0704.二分查找.md

@@ -284,7 +284,6 @@ func search(nums []int, target int) int {
  * @param {number} target
  * @return {number}
  */
-/**
 var search = function(nums, target) {
     let left = 0, right = nums.length - 1;
     // 使用左闭右闭区间
@@ -326,6 +325,46 @@ var search = function(nums, target) {
 };
 ```
 
+**TypeScript**
+
+（版本一）左闭右闭区间
+
+```typescript
+function search(nums: number[], target: number): number {
+    let left: number = 0, right: number = nums.length - 1;
+    while (left <= right) {
+        let mid: number = left + Math.floor((right - left) / 2);
+        if (nums[mid] > target) {
+            right = mid - 1;
+        } else if (nums[mid] < target) {
+            left = mid + 1;
+        } else {
+            return mid;
+        }
+    }
+    return -1;
+};
+```
+
+（版本二）左闭右开区间
+
+```typescript
+function search(nums: number[], target: number): number {
+    let left: number = 0, right: number = nums.length;
+    while (left < right) {
+        let mid: number = left + Math.floor((right - left) / 2);
+        if (nums[mid] > target) {
+            right = mid;
+        } else if (nums[mid] < target) {
+            left = mid + 1;
+        } else {
+            return mid;
+        }
+    }
+    return -1;
+};
+```
+
 **Ruby:**
 
 ```ruby

problems/0739.每日温度.md

@@ -119,7 +119,7 @@ C++代码如下：
 class Solution {
 public:
     vector<int> dailyTemperatures(vector<int>& T) {
-        // 递减栈
+        // 递增栈
         stack<int> st;
         vector<int> result(T.size(), 0);
         st.push(0);
@@ -150,7 +150,7 @@ public:
 class Solution {
 public:
     vector<int> dailyTemperatures(vector<int>& T) {
-        stack<int> st; // 递减栈
+        stack<int> st; // 递增栈
         vector<int> result(T.size(), 0);
         for (int i = 0; i < T.size(); i++) {
             while (!st.empty() && T[i] > T[st.top()]) { // 注意栈不能为空
@@ -178,7 +178,7 @@ public:
 Java：
 ```java
 /**
-     * 单调栈，栈内顺序要么从大到小 要么从小到大,本题从大到笑
+     * 单调栈，栈内顺序要么从大到小 要么从小到大,本题从大到小
      * <p>
      * 入站元素要和当前栈内栈首元素进行比较
      * 若大于栈首则 则与元素下标做差

problems/0746.使用最小花费爬楼梯.md

@@ -82,7 +82,7 @@ dp[1] = cost[1];
 
 **但是稍稍有点难度的动态规划，其遍历顺序并不容易确定下来**。
 
-例如：01背包，都知道两个for循环，一个for遍历物品嵌套一个for遍历背包容量，那么为什么不是一个for遍历背包容量嵌套一个for遍历物品呢？ 以及在使用一维dp数组的时候遍历背包容量为什么要倒叙呢？
+例如：01背包，都知道两个for循环，一个for遍历物品嵌套一个for遍历背包容量，那么为什么不是一个for遍历背包容量嵌套一个for遍历物品呢？ 以及在使用一维dp数组的时候遍历背包容量为什么要倒序呢？
 
 **这些都是遍历顺序息息相关。当然背包问题后续「代码随想录」都会重点讲解的！**
 

problems/0977.有序数组的平方.md

@@ -221,6 +221,35 @@ var sortedSquares = function(nums) {
 };
 ```
 
+Typescript：
+
+双指针法：
+
+```typescript
+function sortedSquares(nums: number[]): number[] {
+    let left: number = 0, right: number = nums.length - 1;
+    let resArr: number[] = new Array(nums.length);
+    let resArrIndex: number = resArr.length - 1;
+    while (left <= right) {
+        if (Math.abs(nums[left]) < Math.abs(nums[right])) {
+            resArr[resArrIndex] = nums[right--] ** 2;
+        } else {
+            resArr[resArrIndex] = nums[left++] ** 2;
+        }
+        resArrIndex--;
+    }
+    return resArr;
+};
+```
+
+骚操作法（暴力思路）：
+
+```typescript
+function sortedSquares(nums: number[]): number[] {
+    return nums.map(i => i * i).sort((a, b) => a - b);
+};
+```
+
 Swift:
 
 ```swift

problems/1002.查找常用字符.md

@@ -58,7 +58,7 @@ words[i] 由小写英文字母组成
 
 先统计第一个字符串所有字符出现的次数，代码如下：
 
-```
+```cpp
 int hash[26] = {0}; // 用来统计所有字符串里字符出现的最小频率
 for (int i = 0; i < A[0].size(); i++) { // 用第一个字符串给hash初始化
     hash[A[0][i] - 'a']++;
@@ -71,7 +71,7 @@ for (int i = 0; i < A[0].size(); i++) { // 用第一个字符串给hash初始化
 
 代码如下：
 
-```
+```cpp
 int hashOtherStr[26] = {0}; // 统计除第一个字符串外字符的出现频率
 for (int i = 1; i < A.size(); i++) {
     memset(hashOtherStr, 0, 26 * sizeof(int));
@@ -84,11 +84,11 @@ for (int i = 1; i < A.size(); i++) {
     }
 }
 ```
-此时hash里统计着字符在所有字符串里出现的最小次数，那么把hash转正题目要求的输出格式就可以了。
+此时hash里统计着字符在所有字符串里出现的最小次数，那么把hash转成题目要求的输出格式就可以了。
 
 代码如下：
 
-```
+```cpp
 // 将hash统计的字符次数，转成输出形式
 for (int i = 0; i < 26; i++) {
     while (hash[i] != 0) { // 注意这里是while，多个重复的字符

problems/1047.删除字符串中的所有相邻重复项.md

@@ -63,7 +63,7 @@
 
 ![1047.删除字符串中的所有相邻重复项](https://code-thinking.cdn.bcebos.com/gifs/1047.删除字符串中的所有相邻重复项.gif)
 
-从栈中弹出剩余元素，此时是字符串ac，因为从栈里弹出的元素是倒叙的，所以在对字符串进行反转一下，就得到了最终的结果。
+从栈中弹出剩余元素，此时是字符串ac，因为从栈里弹出的元素是倒序的，所以在对字符串进行反转一下，就得到了最终的结果。
 
 C++代码 :
 

problems/1049.最后一块石头的重量II.md

@@ -87,7 +87,7 @@ vector<int> dp(15001, 0);
 4. 确定遍历顺序
 
 
-在[动态规划：关于01背包问题，你该了解这些！（滚动数组）](https://programmercarl.com/背包理论基础01背包-2.html)中就已经说明：如果使用一维dp数组，物品遍历的for循环放在外层，遍历背包的for循环放在内层，且内层for循环倒叙遍历！
+在[动态规划：关于01背包问题，你该了解这些！（滚动数组）](https://programmercarl.com/背包理论基础01背包-2.html)中就已经说明：如果使用一维dp数组，物品遍历的for循环放在外层，遍历背包的for循环放在内层，且内层for循环倒序遍历！
 
 代码如下：
 

problems/背包理论基础01背包-2.md

@@ -103,7 +103,7 @@ for(int i = 0; i < weight.size(); i++) { // 遍历物品
 
 为什么呢？
 
-**倒叙遍历是为了保证物品i只被放入一次！**。但如果一旦正序遍历了，那么物品0就会被重复加入多次！
+**倒序遍历是为了保证物品i只被放入一次！**。但如果一旦正序遍历了，那么物品0就会被重复加入多次！
 
 举一个例子：物品0的重量weight[0] = 1，价值value[0] = 15
 
@@ -115,9 +115,9 @@ dp[2] = dp[2 - weight[0]] + value[0] = 30
 
 此时dp[2]就已经是30了，意味着物品0，被放入了两次，所以不能正序遍历。
 
-为什么倒叙遍历，就可以保证物品只放入一次呢？
+为什么倒序遍历，就可以保证物品只放入一次呢？
 
-倒叙就是先算dp[2]
+倒序就是先算dp[2]
 
 dp[2] = dp[2 - weight[0]] + value[0] = 15  （dp数组已经都初始化为0）
 
@@ -125,7 +125,7 @@ dp[1] = dp[1 - weight[0]] + value[0] = 15
 
 所以从后往前循环，每次取得状态不会和之前取得状态重合，这样每种物品就只取一次了。
 
-**那么问题又来了，为什么二维dp数组历的时候不用倒叙呢？**
+**那么问题又来了，为什么二维dp数组历的时候不用倒序呢？**
 
 因为对于二维dp，dp[i][j]都是通过上一层即dp[i - 1][j]计算而来，本层的dp[i][j]并不会被覆盖！
 

problems/链表理论基础.md

@@ -168,6 +168,32 @@ public class ListNode {
 }
 ```
 
+JavaScript:
+
+```javascript
+class ListNode {
+  val;
+  next = null;
+  constructor(value) {
+    this.val = value;
+    this.next = null;
+  }
+}
+```
+
+TypeScript:
+
+```typescript
+class ListNode {
+  public val: number;
+  public next: ListNode = null;
+  constructor(value: number) {
+    this.val = value;
+    this.next = null;
+  }
+}
+```
+
 Python：