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

problems/0037.解数独.md

@@ -439,6 +439,55 @@ var solveSudoku = function(board) {
 };
 ```
 
+### TypeScript
+
+```typescript
+/**
+ Do not return anything, modify board in-place instead.
+ */
+function isValid(col: number, row: number, val: string, board: string[][]): boolean {
+    let n: number = board.length;
+    // 列向检查
+    for (let rowIndex = 0; rowIndex < n; rowIndex++) {
+        if (board[rowIndex][col] === val) return false;
+    }
+    // 横向检查
+    for (let colIndex = 0; colIndex < n; colIndex++) {
+        if (board[row][colIndex] === val) return false;
+    }
+    // 九宫格检查
+    const startX = Math.floor(col / 3) * 3;
+    const startY = Math.floor(row / 3) * 3;
+    for (let rowIndex = startY; rowIndex < startY + 3; rowIndex++) {
+        for (let colIndex = startX; colIndex < startX + 3; colIndex++) {
+            if (board[rowIndex][colIndex] === val) return false;
+        }
+    }
+    return true;
+}
+function solveSudoku(board: string[][]): void {
+    let n: number = 9;
+    backTracking(n, board);
+    function backTracking(n: number, board: string[][]): boolean {
+        for (let row = 0; row < n; row++) {
+            for (let col = 0; col < n; col++) {
+                if (board[row][col] === '.') {
+                    for (let i = 1; i <= n; i++) {
+                        if (isValid(col, row, String(i), board)) {
+                            board[row][col] = String(i);
+                            if (backTracking(n, board) === true) return true;
+                            board[row][col] = '.';
+                        }
+                    }
+                    return false;
+                }
+            }
+        }
+        return true;
+    }
+};
+```
+
 ### C 
 
 ```C

problems/0046.全排列.md

@@ -331,6 +331,34 @@ var permute = function(nums) {
 
 ```
 
+## TypeScript
+
+```typescript
+function permute(nums: number[]): number[][] {
+    const resArr: number[][] = [];
+    const helperSet: Set<number> = new Set();
+    backTracking(nums, []);
+    return resArr;
+    function backTracking(nums: number[], route: number[]): void {
+        if (route.length === nums.length) {
+            resArr.push(route.slice());
+            return;
+        }
+        let tempVal: number;
+        for (let i = 0, length = nums.length; i < length; i++) {
+            tempVal = nums[i];
+            if (!helperSet.has(tempVal)) {
+                route.push(tempVal);
+                helperSet.add(tempVal);
+                backTracking(nums, route);
+                route.pop();
+                helperSet.delete(tempVal);
+            }
+        }
+    }
+};
+```
+
 ### C
 
 ```c

problems/0047.全排列II.md

@@ -292,6 +292,34 @@ var permuteUnique = function (nums) {
 
 ```
 
+### TypeScript
+
+```typescript
+function permuteUnique(nums: number[]): number[][] {
+    nums.sort((a, b) => a - b);
+    const resArr: number[][] = [];
+    const usedArr: boolean[] = new Array(nums.length).fill(false);
+    backTracking(nums, []);
+    return resArr;
+    function backTracking(nums: number[], route: number[]): void {
+        if (route.length === nums.length) {
+            resArr.push(route.slice());
+            return;
+        }
+        for (let i = 0, length = nums.length; i < length; i++) {
+            if (i > 0 && nums[i] === nums[i - 1] && usedArr[i - 1] === false) continue;
+            if (usedArr[i] === false) {
+                route.push(nums[i]);
+                usedArr[i] = true;
+                backTracking(nums, route);
+                usedArr[i] = false;
+                route.pop();
+            }
+        }
+    }
+};
+```
+
 ### Swift
 
 ```swift

problems/0051.N皇后.md

@@ -457,6 +457,58 @@ var solveNQueens = function(n) {
 };
 ```
 
+## TypeScript
+
+```typescript
+function solveNQueens(n: number): string[][] {
+    const board: string[][] = new Array(n).fill(0).map(_ => new Array(n).fill('.'));
+    const resArr: string[][] = [];
+    backTracking(n, 0, board);
+    return resArr;
+    function backTracking(n: number, rowNum: number, board: string[][]): void {
+        if (rowNum === n) {
+            resArr.push(transformBoard(board));
+            return;
+        }
+        for (let i = 0; i < n; i++) {
+            if (isValid(i, rowNum, board) === true) {
+                board[rowNum][i] = 'Q';
+                backTracking(n, rowNum + 1, board);
+                board[rowNum][i] = '.';
+            }
+        }
+    }
+};
+function isValid(col: number, row: number, board: string[][]): boolean {
+    const n: number = board.length;
+    if (col < 0 || col >= n || row < 0 || row >= n) return false;
+    // 检查列
+    for (let row of board) {
+        if (row[col] === 'Q') return false;
+    }
+    // 检查45度方向
+    let x: number = col,
+        y: number = row;
+    while (y >= 0 && x < n) {
+        if (board[y--][x++] === 'Q') return false;
+    }
+    // 检查135度方向
+    x = col;
+    y = row;
+    while (x >= 0 && y >= 0) {
+        if (board[y--][x--] === 'Q') return false;
+    }
+    return true;
+}
+function transformBoard(board: string[][]): string[] {
+    const resArr = [];
+    for (let row of board) {
+        resArr.push(row.join(''));
+    }
+    return resArr;
+}
+```
+
 ### Swift
 
 ```swift

problems/0053.最大子序和.md

@@ -230,6 +230,41 @@ var maxSubArray = function(nums) {
 };
 ```
 
+### TypeScript
+
+**贪心**
+
+```typescript
+function maxSubArray(nums: number[]): number {
+    let curSum: number = 0;
+    let resMax: number = -Infinity;
+    for (let i = 0, length = nums.length; i < length; i++) {
+        curSum += nums[i];
+        resMax = Math.max(curSum, resMax);
+        if (curSum < 0) curSum = 0;
+    }
+    return resMax;
+};
+```
+
+**动态规划**
+
+```typescript
+// 动态规划
+function maxSubArray(nums: number[]): number {
+    const length = nums.length;
+    if (length === 0) return 0;
+    const dp: number[] = [];
+    dp[0] = nums[0];
+    let resMax: number = nums[0];
+    for (let i = 1; i < length; i++) {
+        dp[i] = Math.max(dp[i - 1] + nums[i], nums[i]);
+        resMax = Math.max(resMax, dp[i]);
+    }
+    return resMax;
+};
+```
+
 
 
 -----------------------

problems/0098.验证二叉搜索树.md

@@ -408,6 +408,27 @@ class Solution:
         return True
 
 ```
+```python
+# 遵循Carl的写法，只添加了节点判断的部分
+class Solution:
+    def isValidBST(self, root: TreeNode) -> bool:
+        # method 2
+        que, pre = [], None
+        while root or que:
+            while root:
+                que.append(root)
+                root = root.left
+            root = que.pop()
+            # 对第一个节点只做记录，对后面的节点进行比较
+            if pre is None:
+                pre = root.val
+            else:
+                if pre >= root.val: return False
+                pre = root.val
+            root = root.right
+        return True
+```
+
 ## Go
 
 ```Go

problems/0101.对称二叉树.md

@@ -437,41 +437,6 @@ class Solution:
         return True
 ```
 
-层序遍历
-
-```python
-class Solution:
-    def isSymmetric(self, root: TreeNode) -> bool:
-        if not root: return True
-        que, cnt = [[root.left, root.right]], 1
-        while que:
-            nodes, tmp, sign = que.pop(), [], False
-            for node in nodes:
-                if not node:
-                    tmp.append(None)
-                    tmp.append(None)
-                else:
-                    if node.left:
-                        tmp.append(node.left)
-                        sign = True
-                    else:
-                        tmp.append(None)
-                    if node.right:
-                        tmp.append(node.right)
-                        sign = True
-                    else:
-                        tmp.append(None)
-            p1, p2 = 0, len(nodes) - 1
-            while p1 < p2:
-                if (not nodes[p1] and nodes[p2]) or (nodes[p1] and not nodes[p2]): return False
-                elif nodes[p1] and nodes[p2] and nodes[p1].val != nodes[p2].val: return False
-                p1 += 1
-                p2 -= 1
-            if sign: que.append(tmp)
-            cnt += 1
-        return True
-```
-
 ## Go
 
 ```go

problems/0106.从中序与后序遍历序列构造二叉树.md

@@ -790,7 +790,7 @@ func findRootIndex(target int,inorder []int) int{
 
 ```javascript
 var buildTree = function(inorder, postorder) {
-    if (!preorder.length) return null;
+    if (!inorder.length) return null;
     const rootVal = postorder.pop(); // 从后序遍历的数组中获取中间节点的值， 即数组最后一个值
     let rootIndex = inorder.indexOf(rootVal); // 获取中间节点在中序遍历中的下标
     const root = new TreeNode(rootVal); // 创建中间节点

problems/0108.将有序数组转换为二叉搜索树.md

@@ -355,6 +355,7 @@ func sortedArrayToBST(nums []int) *TreeNode {
 ```
 
 ## JavaScript 
+递归
 
 ```javascript
 var sortedArrayToBST = function (nums) {
@@ -372,7 +373,44 @@ var sortedArrayToBST = function (nums) {
     return buildTree(nums, 0, nums.length - 1);
 };
 ```
+迭代
+```JavaScript
+var sortedArrayToBST = function(nums) {
+    if(nums.length===0){
+        return null;
+    }
+    let root=new TreeNode(0); //初始根节点
+    let nodeQue=[root]; //放遍历的节点,并初始化
+    let leftQue=[0]; //放左区间的下标,初始化
+    let rightQue=[nums.length-1]; // 放右区间的下标
     
+    while(nodeQue.length){
+        let curNode=nodeQue.pop();
+        let left=leftQue.pop();
+        let right=rightQue.pop();
+        let mid=left+Math.floor((right-left)/2);
+        
+        curNode.val=nums[mid]; //将下标为mid的元素给中间节点
+        
+//         处理左区间
+        if(left<=mid-1){
+            curNode.left=new TreeNode(0);
+            nodeQue.push(curNode.left);
+            leftQue.push(left);
+            rightQue.push(mid-1);
+        }
+        
+//         处理右区间
+        if(right>=mid+1){
+            curNode.right=new TreeNode(0);
+            nodeQue.push(curNode.right);
+            leftQue.push(mid+1);
+            rightQue.push(right);
+        }
+    }
+    return root;
+};
+```
 ## TypeScript
 
 ```typescript

problems/0122.买卖股票的最佳时机II.md

@@ -40,7 +40,7 @@
 本题首先要清楚两点：
 
 * 只有一只股票！
-* 当前只有买股票或者买股票的操作
+* 当前只有买股票或者卖股票的操作
 
 想获得利润至少要两天为一个交易单元。
 

problems/0202.快乐数.md

@@ -315,5 +315,75 @@ class Solution {
 }
 ```
 
+C:
+```C
+typedef struct HashNodeTag {
+    int key;   /* num  */
+    struct HashNodeTag *next;
+}HashNode;
+
+/* Calcualte the hash key */
+static inline int hash(int key, int size) {
+    int index = key % size;
+    return (index > 0) ? (index) : (-index);
+}
+
+/* Calculate the sum of the squares of its digits*/
+static inline int calcSquareSum(int num) {
+    unsigned int sum = 0;
+    while(num > 0) {
+        sum += (num % 10) * (num % 10);
+        num = num/10; 
+    }
+    return sum;
+}
+
+#define HASH_TABLE_SIZE     (32)
+
+bool isHappy(int n){
+    int sum = n;
+    int index = 0;
+    bool bHappy = false;
+    bool bExit = false;
+    /* allocate the memory for hash table with chaining method*/
+    HashNode ** hashTable = (HashNode **)calloc(HASH_TABLE_SIZE, sizeof(HashNode));
+    
+    while(bExit == false) {
+        /* check if n has been calculated */
+        index = hash(n, HASH_TABLE_SIZE);
+
+        HashNode ** p = hashTable + index;
+
+        while((*p) && (bExit == false)) {
+            /* Check if this num was calculated, if yes, this will be endless loop */
+            if((*p)->key == n) { 
+                bHappy = false;
+                bExit = true;
+            }
+            /* move to next node of the same index */
+            p = &((*p)->next);
+        }
+
+        /* put n intot hash table */
+        HashNode * newNode = (HashNode *)malloc(sizeof(HashNode));
+        newNode->key = n;
+        newNode->next = NULL;
+
+        *p = newNode;
+
+        sum = calcSquareSum(n);
+        if(sum == 1) {
+            bHappy = true;
+            bExit = true;
+        }
+        else {
+            n = sum;
+            
+        }
+    }
+
+    return bHappy;
+}
+```
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0232.用栈实现队列.md

@@ -275,15 +275,11 @@ func (this *MyQueue) Pop() int {
 
 /** Get the front element. */
 func (this *MyQueue) Peek() int {
-    for len(this.stack) != 0 {
+    val := this.Pop()
-        val := this.stack[len(this.stack)-1]
+    if val == 0 {
-        this.stack = this.stack[:len(this.stack)-1]
-        this.back = append(this.back, val)
-    }
-    if len(this.back) == 0 {
         return 0
     }
-    val := this.back[len(this.back)-1]
+    this.back = append(this.back, val)
     return val
 }
 

problems/0332.重新安排行程.md

@@ -506,6 +506,50 @@ var findItinerary = function(tickets) {
 
 ```
 
+### TypeScript
+
+```typescript
+function findItinerary(tickets: string[][]): string[] {
+    /**
+        TicketsMap 实例：
+        { NRT: Map(1) { 'JFK' => 1 }, JFK: Map(2) { 'KUL' => 1, 'NRT' => 1 } }
+        这里选择Map数据结构的原因是：与Object类型的一个主要差异是，Map实例会维护键值对的插入顺序。
+     */
+    type TicketsMap = {
+        [index: string]: Map<string, number>
+    };
+    tickets.sort((a, b) => {
+        return a[1] < b[1] ? -1 : 1;
+    });
+    const ticketMap: TicketsMap = {};
+    for (const [from, to] of tickets) {
+        if (ticketMap[from] === undefined) {
+            ticketMap[from] = new Map();
+        }
+        ticketMap[from].set(to, (ticketMap[from].get(to) || 0) + 1);
+    }
+    const resRoute = ['JFK'];
+    backTracking(tickets.length, ticketMap, resRoute);
+    return resRoute;
+    function backTracking(ticketNum: number, ticketMap: TicketsMap, route: string[]): boolean {
+        if (route.length === ticketNum + 1) return true;
+        const targetMap = ticketMap[route[route.length - 1]];
+        if (targetMap !== undefined) {
+            for (const [to, count] of targetMap.entries()) {
+                if (count > 0) {
+                    route.push(to);
+                    targetMap.set(to, count - 1);
+                    if (backTracking(ticketNum, ticketMap, route) === true) return true;
+                    targetMap.set(to, count);
+                    route.pop();
+                }
+            }
+        }
+        return false;
+    }
+};
+```
+
 ### Swift
 
 直接迭代tickets数组：

problems/0349.两个数组的交集.md

@@ -281,6 +281,38 @@ impl Solution {
     }
 }
 ```
+
+C:
+```C
+int* intersection1(int* nums1, int nums1Size, int* nums2, int nums2Size, int* returnSize){
+
+    int nums1Cnt[1000] = {0};
+    int lessSize = nums1Size < nums2Size ? nums1Size : nums2Size;
+    int * result = (int *) calloc(lessSize, sizeof(int));
+    int resultIndex = 0;
+    int* tempNums;
+    
+    int i;
+
+    /* Calculate the number's counts for nums1 array */
+    for(i = 0; i < nums1Size; i ++) {
+        nums1Cnt[nums1[i]]++;
+    }
+
+    /* Check if the value in nums2 is existing in nums1 count array */
+    for(i = 0; i < nums2Size; i ++) {
+        if(nums1Cnt[nums2[i]] > 0) {
+            result[resultIndex] = nums2[i];
+            resultIndex ++;
+            /* Clear this count to avoid duplicated value */
+            nums1Cnt[nums2[i]] = 0;
+        }
+    }
+    * returnSize = resultIndex;
+    return result;
+}
+```
+
 ## 相关题目
 
 * 350.两个数组的交集 II

problems/0376.摆动序列.md

@@ -298,5 +298,55 @@ var wiggleMaxLength = function(nums) {
 };
 ```
 
+### TypeScript
+
+**贪心**
+
+```typescript
+function wiggleMaxLength(nums: number[]): number {
+    let length: number = nums.length;
+    if (length <= 1) return length;
+    let preDiff: number = 0;
+    let curDiff: number = 0;
+    let count: number = 1;
+    for (let i = 1; i < length; i++) {
+        curDiff = nums[i] - nums[i - 1];
+        if (
+            (preDiff <= 0 && curDiff > 0) ||
+            (preDiff >= 0 && curDiff < 0)
+        ) {
+            preDiff = curDiff;
+            count++;
+        }
+    }
+    return count;
+};
+```
+
+**动态规划**
+
+```typescript
+function wiggleMaxLength(nums: number[]): number {
+    const length: number = nums.length;
+    if (length <= 1) return length;
+    const dp: number[][] = new Array(length).fill(0).map(_ => []);
+    dp[0][0] = 1;   // 第一个数作为波峰
+    dp[0][1] = 1;   // 第一个数作为波谷
+    for (let i = 1; i < length; i++) {
+        dp[i][0] = 1;
+        dp[i][1] = 1;
+        for (let j = 0; j < i; j++) {
+            if (nums[j] < nums[i]) dp[i][0] = Math.max(dp[i][0], dp[j][1] + 1);
+        }
+        for (let j = 0; j < i; j++) {
+            if (nums[j] > nums[i]) dp[i][1] = Math.max(dp[i][1], dp[j][0] + 1);
+        }
+    }
+    return Math.max(dp[length - 1][0], dp[length - 1][1]);
+};
+```
+
+
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0455.分发饼干.md

@@ -209,7 +209,50 @@ var findContentChildren = function(g, s) {
 
 ```
 
+### TypeScript
+
+```typescript
+// 大饼干尽量喂胃口大的
+function findContentChildren(g: number[], s: number[]): number {
+    g.sort((a, b) => a - b);
+    s.sort((a, b) => a - b);
+    const childLength: number = g.length,
+        cookieLength: number = s.length;
+    let curChild: number = childLength - 1,
+        curCookie: number = cookieLength - 1;
+    let resCount: number = 0;
+    while (curChild >= 0 && curCookie >= 0) {
+        if (g[curChild] <= s[curCookie]) {
+            curCookie--;
+            resCount++;
+        }
+        curChild--;
+    }
+    return resCount;
+};
+```
+
+```typescript
+// 小饼干先喂饱小胃口的
+function findContentChildren(g: number[], s: number[]): number {
+    g.sort((a, b) => a - b);
+    s.sort((a, b) => a - b);
+    const childLength: number = g.length,
+        cookieLength: number = s.length;
+    let curChild: number = 0,
+        curCookie: number = 0;
+    while (curChild < childLength && curCookie < cookieLength) {
+        if (g[curChild] <= s[curCookie]) {
+            curChild++;
+        }
+        curCookie++;
+    }
+    return curChild;
+};
+```
+
 ### C 
+
 ```c
 int cmp(int* a, int* b) {
     return *a - *b;

problems/0491.递增子序列.md

@@ -396,7 +396,35 @@ var findSubsequences = function(nums) {
 
 ```
 
+## TypeScript
+
+```typescript
+function findSubsequences(nums: number[]): number[][] {
+    const resArr: number[][] = [];
+    backTracking(nums, 0, []);
+    return resArr;
+    function backTracking(nums: number[], startIndex: number, route: number[]): void {
+        let length: number = nums.length;
+        if (route.length >= 2) {
+            resArr.push(route.slice());
+        }
+        const usedSet: Set<number> = new Set();
+        for (let i = startIndex; i < length; i++) {
+            if (
+                nums[i] < route[route.length - 1] ||
+                usedSet.has(nums[i])
+            ) continue;
+            usedSet.add(nums[i]);
+            route.push(nums[i]);
+            backTracking(nums, i + 1, route);
+            route.pop();
+        }
+    }
+};
+```
+
 ### C 
+
 ```c
 int* path;
 int pathTop;

problems/0704.二分查找.md

@@ -276,7 +276,7 @@ func search(nums []int, target int) int {
 ```
 
 **JavaScript:**
-（版本一）左闭右闭区间
+（版本一）左闭右闭区间 [left, right]
 
 ```js
 /**
@@ -285,10 +285,12 @@ func search(nums []int, target int) int {
  * @return {number}
  */
 var search = function(nums, target) {
+    // right是数组最后一个数的下标，num[right]在查找范围内，是左闭右闭区间
     let left = 0, right = nums.length - 1;
-    // 使用左闭右闭区间
+    // 当left=right时，由于nums[right]在查找范围内，所以要包括此情况
     while (left <= right) {
         let mid = left + Math.floor((right - left)/2);
+        // 如果中间数大于目标值，要把中间数排除查找范围，所以右边界更新为mid-1；如果右边界更新为mid，那中间数还在下次查找范围内
         if (nums[mid] > target) {
             right = mid - 1;  // 去左面闭区间寻找
         } else if (nums[mid] < target) {
@@ -300,7 +302,7 @@ var search = function(nums, target) {
     return -1;
 };
 ```
-（版本二）左闭右开区间
+（版本二）左闭右开区间 [left, right)
 
 ```js
 /**
@@ -309,10 +311,13 @@ var search = function(nums, target) {
  * @return {number}
  */
 var search = function(nums, target) {
+    // right是数组最后一个数的下标+1，nums[right]不在查找范围内，是左闭右开区间
     let left = 0, right = nums.length;    
-    // 使用左闭右开区间 [left, right)
+    // 当left=right时，由于nums[right]不在查找范围，所以不必包括此情况
     while (left < right) {
         let mid = left + Math.floor((right - left)/2);
+        // 如果中间值大于目标值，中间值不应在下次查找的范围内，但中间值的前一个值应在；
+        // 由于right本来就不在查找范围内，所以将右边界更新为中间值，如果更新右边界为mid-1则将中间值的前一个值也踢出了下次寻找范围
         if (nums[mid] > target) {
             right = mid;  // 去左区间寻找
         } else if (nums[mid] < target) {

problems/前序/ACM模式如何构建二叉树.md

@@ -220,6 +220,73 @@ int main() {
 ## Go 
 
 ```Go
+package main
+
+import "fmt"
+
+type TreeNode struct {
+    Val   int
+    Left  *TreeNode
+    Right *TreeNode
+}
+
+func constructBinaryTree(array []int) *TreeNode {
+    var root *TreeNode
+    nodes := make([]*TreeNode, len(array))
+
+    // 初始化二叉树节点
+    for i := 0; i < len(nodes); i++ {
+	var node *TreeNode
+	if array[i] != -1 {
+	    node = &TreeNode{Val: array[i]}
+	}
+	nodes[i] = node
+	if i == 0 {
+	    root = node
+	}
+    }
+    // 串联节点
+    for i := 0; i*2+2 < len(array); i++ {
+        if nodes[i] != nil {
+	    nodes[i].Left = nodes[i*2+1]
+	    nodes[i].Right = nodes[i*2+2]
+	}
+    }
+    return root
+}
+
+func printBinaryTree(root *TreeNode, n int) {
+    var queue []*TreeNode
+    if root != nil {
+	queue = append(queue, root)
+    }
+
+    result := []int{}
+    for len(queue) > 0 {
+	for j := 0; j < len(queue); j++ {
+	    node := queue[j]
+	    if node != nil {
+		result = append(result, node.Val)
+		queue = append(queue, node.Left)
+		queue = append(queue, node.Right)
+	    } else {
+	        result = append(result, -1)
+	    }
+        }
+	// 清除队列中的本层节点, 进入下一层遍历
+	queue = queue[len(queue):]
+    }
+    
+    // 参数n控制输出值数量, 否则二叉树最后一层叶子节点的孩子节点也会被打印(但是这些孩子节点是不存在的).
+    fmt.Println(result[:n])
+}
+
+func main() {
+    array := []int{4, 1, 6, 0, 2, 5, 7, -1, -1, -1, 3, -1, -1, -1, 8}
+    root := constructBinaryTree(array)
+    printBinaryTree(root, len(array))
+}
+
 ```
 
 ## JavaScript

problems/周总结/20201112回溯周末总结.md

@@ -76,7 +76,7 @@
 * 空间复杂度：$O(n)$，递归深度为n，所以系统栈所用空间为$O(n)$，每一层递归所用的空间都是常数级别，注意代码里的result和path都是全局变量，就算是放在参数里，传的也是引用，并不会新申请内存空间，最终空间复杂度为$O(n)$。
 
 排列问题分析：
-* 时间复杂度：$O(n!)$，这个可以从排列的树形图中很明显发现，每一层节点为n，第二层每一个分支都延伸了n-1个分支，再往下又是n-2个分支，所以一直到叶子节点一共就是 n * n-1 * n-2 * ..... 1 = n!。
+* 时间复杂度：$O(n!)$，这个可以从排列的树形图中很明显发现，每一层节点为n，第二层每一个分支都延伸了n-1个分支，再往下又是n-2个分支，所以一直到叶子节点一共就是 n * n-1 * n-2 * ..... 1 = n!。每个叶子节点都会有一个构造全排列填进数组的操作（对应的代码：`result.push_back(path)`），该操作的复杂度为$O(n)$。所以，最终时间复杂度为：n * n!，简化为$O(n!)$。
 * 空间复杂度：$O(n)$，和子集问题同理。
 
 组合问题分析：

problems/回溯算法去重问题的另一种写法.md

@@ -365,6 +365,87 @@ class Solution:
         return res
 ```
 
+TypeScript：
+
+**90.子集II**
+
+```typescript
+function subsetsWithDup(nums: number[]): number[][] {
+    nums.sort((a, b) => a - b);
+    const resArr: number[][] = [];
+    backTraking(nums, 0, []);
+    return resArr;
+    function backTraking(nums: number[], startIndex: number, route: number[]): void {
+        resArr.push(route.slice());
+        const helperSet: Set<number> = new Set();
+        for (let i = startIndex, length = nums.length; i < length; i++) {
+            if (helperSet.has(nums[i])) continue;
+            helperSet.add(nums[i]);
+            route.push(nums[i]);
+            backTraking(nums, i + 1, route);
+            route.pop();
+        }
+    }
+};
+```
+
+**40. 组合总和 II**
+
+```typescript
+function combinationSum2(candidates: number[], target: number): number[][] {
+    candidates.sort((a, b) => a - b);
+    const resArr: number[][] = [];
+    backTracking(candidates, target, 0, 0, []);
+    return resArr;
+    function backTracking(
+        candidates: number[], target: number,
+        curSum: number, startIndex: number, route: number[]
+    ) {
+        if (curSum > target) return;
+        if (curSum === target) {
+            resArr.push(route.slice());
+            return;
+        }
+        const helperSet: Set<number> = new Set();
+        for (let i = startIndex, length = candidates.length; i < length; i++) {
+            let tempVal: number = candidates[i];
+            if (helperSet.has(tempVal)) continue;
+            helperSet.add(tempVal);
+            route.push(tempVal);
+            backTracking(candidates, target, curSum + tempVal, i + 1, route);
+            route.pop();
+
+        }
+    }
+};
+```
+
+**47. 全排列 II**
+
+```typescript
+function permuteUnique(nums: number[]): number[][] {
+    const resArr: number[][] = [];
+    const usedArr: boolean[] = [];
+    backTracking(nums, []);
+    return resArr;
+    function backTracking(nums: number[], route: number[]): void {
+        if (nums.length === route.length) {
+            resArr.push(route.slice());
+            return;
+        }
+        const usedSet: Set<number> = new Set();
+        for (let i = 0, length = nums.length; i < length; i++) {
+            if (usedArr[i] === true || usedSet.has(nums[i])) continue;
+            usedSet.add(nums[i]);
+            route.push(nums[i]);
+            usedArr[i] = true;
+            backTracking(nums, route);
+            usedArr[i] = false;
+            route.pop();
+        }
+    }
+};
+```
 
 Go：