Merge branch 'youngyangyang04:master' into master

problems/0019.删除链表的倒数第N个节点.md

@@ -181,7 +181,71 @@ var removeNthFromEnd = function(head, n) {
     return ret.next;
 };
 ```
+TypeScript:
+
+版本一（快慢指针法）：
+
+```typescript
+function removeNthFromEnd(head: ListNode | null, n: number): ListNode | null {
+    let newHead: ListNode | null = new ListNode(0, head);
+    let slowNode: ListNode | null = newHead,
+        fastNode: ListNode | null = newHead;
+    for (let i = 0; i < n; i++) {
+        fastNode = fastNode.next;
+    }
+    while (fastNode.next) {
+        fastNode = fastNode.next;
+        slowNode = slowNode.next;
+    }
+    slowNode.next = slowNode.next.next;
+    return newHead.next;
+};
+```
+
+版本二（计算节点总数法）：
+
+```typescript
+function removeNthFromEnd(head: ListNode | null, n: number): ListNode | null {
+    let curNode: ListNode | null = head;
+    let listSize: number = 0;
+    while (curNode) {
+        curNode = curNode.next;
+        listSize++;
+    }
+    if (listSize === n) {
+        head = head.next;
+    } else {
+        curNode = head;
+        for (let i = 0; i < listSize - n - 1; i++) {
+            curNode = curNode.next;
+        }
+        curNode.next = curNode.next.next;
+    }
+    return head;
+};
+```
+
+版本三（递归倒退n法）：
+
+```typescript
+function removeNthFromEnd(head: ListNode | null, n: number): ListNode | null {
+    let newHead: ListNode | null = new ListNode(0, head);
+    let cnt = 0;
+    function recur(node) {
+        if (node === null) return;
+        recur(node.next);
+        cnt++;
+        if (cnt === n + 1) {
+            node.next = node.next.next;
+        }
+    }
+    recur(newHead);
+    return newHead.next;
+};
+```
+
 Kotlin:
+
 ```Kotlin
 fun removeNthFromEnd(head: ListNode?, n: Int): ListNode? {
     val pre = ListNode(0).apply {

problems/0024.两两交换链表中的节点.md

@@ -250,6 +250,38 @@ var swapPairs = function (head) {
 };
 ```
 
+TypeScript：
+
+```typescript
+function swapPairs(head: ListNode | null): ListNode | null {
+    /**
+     * 初始状态：
+     * curNode -> node1 -> node2 -> tmepNode
+     * 转换过程：
+     * curNode -> node2
+     * curNode -> node2 -> node1
+     * curNode -> node2 -> node1 -> tempNode
+     * curNode = node1
+     */
+    let retNode: ListNode | null = new ListNode(0, head),
+        curNode: ListNode | null = retNode,
+        node1: ListNode | null = null,
+        node2: ListNode | null = null,
+        tempNode: ListNode | null = null;
+
+    while (curNode && curNode.next && curNode.next.next) {
+        node1 = curNode.next;
+        node2 = curNode.next.next;
+        tempNode = node2.next;
+        curNode.next = node2;
+        node2.next = node1;
+        node1.next = tempNode;
+        curNode = node1;
+    }
+    return retNode.next;
+};
+```
+
 Kotlin:
 
 ```kotlin

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

@@ -389,6 +389,51 @@ class Solution:
 ### Go
 
 ```go
+func searchRange(nums []int, target int) []int {
+    leftBorder := getLeft(nums, target)
+    rightBorder := getRight(nums, target)
+    // 情况一
+    if leftBorder == -2 || rightBorder == -2 {
+        return []int{-1, -1}
+    }
+    // 情况三
+    if rightBorder - leftBorder > 1 {
+        return []int{leftBorder + 1, rightBorder - 1}
+    }
+    // 情况二
+    return []int{-1, -1}
+}
+
+func getLeft(nums []int, target int) int {
+    left, right := 0, len(nums)-1
+    border := -2 // 记录border没有被赋值的情况；这里不能赋值-1，target = num[0]时，会无法区分情况一和情况二
+    for left <= right { // []闭区间
+	mid := left + ((right - left) >> 1)
+	if nums[mid] >= target { // 找到第一个等于target的位置
+	    right = mid - 1
+            border = right
+	} else {
+	    left =  mid + 1
+	}
+    }
+    return border
+}
+
+func getRight(nums []int, target int) int {
+    left, right := 0, len(nums) - 1
+    border := -2
+    for left <= right {
+	mid := left + ((right - left) >> 1)
+	if nums[mid] > target { 
+		right = mid - 1
+	} else { // 找到第一个大于target的位置
+	    left = mid + 1
+            border = left
+	}
+    }
+    return border
+
+}
 ```
 
 ### JavaScript

problems/0059.螺旋矩阵II.md

@@ -287,6 +287,51 @@ var generateMatrix = function(n) {
 
 ```
 
+TypeScript:
+
+```typescript
+function generateMatrix(n: number): number[][] {
+    let loopNum: number = Math.floor(n / 2);
+    const resArr: number[][] = new Array(n).fill(1).map(i => new Array(n));
+    let chunkNum: number = n - 1;
+    let startX: number = 0;
+    let startY: number = 0;
+    let value: number = 1;
+    let x: number, y: number;
+    while (loopNum--) {
+        x = startX;
+        y = startY;
+        while (x < startX + chunkNum) {
+            resArr[y][x] = value;
+            x++;
+            value++;
+        }
+        while (y < startY + chunkNum) {
+            resArr[y][x] = value;
+            y++;
+            value++;
+        }
+        while (x > startX) {
+            resArr[y][x] = value;
+            x--;
+            value++;
+        }
+        while (y > startY) {
+            resArr[y][x] = value;
+            y--;
+            value++;
+        }
+        startX++;
+        startY++;
+        chunkNum -= 2;
+    }
+    if (n % 2 === 1) {
+        resArr[startX][startY] = value;
+    }
+    return resArr;
+};
+```
+
 Go:
 
 ```go

problems/0062.不同路径.md

@@ -347,6 +347,42 @@ var uniquePaths = function(m, n) {
 };
 ```
 
+### C
+```c
+//初始化dp数组
+int **initDP(int m, int n) {
+    //动态开辟dp数组
+    int **dp = (int**)malloc(sizeof(int *) * m);
+    int i, j;
+    for(i = 0; i < m; ++i) {
+        dp[i] = (int *)malloc(sizeof(int) * n);
+    }
+
+    //从0，0到i,0只有一种走法，所以dp[i][0]都是1，同理dp[0][j]也是1
+    for(i = 0; i < m; ++i) 
+        dp[i][0] = 1;
+    for(j = 0; j < n; ++j) 
+        dp[0][j] = 1;
+    return dp;
+}
+
+int uniquePaths(int m, int n){
+    //dp数组，dp[i][j]代表从dp[0][0]到dp[i][j]有几种走法
+    int **dp = initDP(m, n);
+
+    int i, j;
+    //到达dp[i][j]只能从dp[i-1][j]和dp[i][j-1]出发
+    //dp[i][j] = dp[i-1][j] + dp[i][j-1]
+    for(i = 1; i < m; ++i) {
+        for(j = 1; j < n; ++j) {
+            dp[i][j] = dp[i-1][j] + dp[i][j-1];
+        }
+    }
+    int result = dp[m-1][n-1];
+    free(dp);
+    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/0063.不同路径II.md

@@ -333,6 +333,60 @@ var uniquePathsWithObstacles = function(obstacleGrid) {
 };
 ```
 
+C
+```c
+//初始化dp数组
+int **initDP(int m, int n, int** obstacleGrid) {
+    int **dp = (int**)malloc(sizeof(int*) * m);
+    int i, j;
+    //初始化每一行数组
+    for(i = 0; i < m; ++i) {
+        dp[i] = (int*)malloc(sizeof(int) * n);
+    }
+
+    //先将第一行第一列设为0
+    for(i = 0; i < m; ++i) {
+        dp[i][0] = 0;
+    }
+    for(j = 0; j < n; ++j) {
+        dp[0][j] = 0;
+    }
+
+    //若碰到障碍，之后的都走不了。退出循环
+    for(i = 0; i < m; ++i) {
+        if(obstacleGrid[i][0]) {
+            break;
+        }
+        dp[i][0] = 1;
+    }
+    for(j = 0; j < n; ++j) {
+        if(obstacleGrid[0][j])
+            break;
+        dp[0][j] = 1;
+    }
+    return dp;
+}
+
+int uniquePathsWithObstacles(int** obstacleGrid, int obstacleGridSize, int* obstacleGridColSize){
+    int m = obstacleGridSize, n = *obstacleGridColSize;
+    //初始化dp数组
+    int **dp = initDP(m, n, obstacleGrid);
+
+    int i, j;
+    for(i = 1; i < m; ++i) {
+        for(j = 1; j < n; ++j) {
+            //若当前i,j位置有障碍
+            if(obstacleGrid[i][j])
+                //路线不同
+                dp[i][j] = 0;
+            else
+                dp[i][j] = dp[i-1][j] + dp[i][j-1];
+        }
+    }
+    //返回最后终点的路径个数
+    return 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

@@ -296,6 +296,48 @@ var climbStairs = function(n) {
 };
 ```
 
+### C
+```c
+int climbStairs(int n){
+    //若n<=2，返回n
+    if(n <= 2)
+        return n;
+    //初始化dp数组，数组大小为n+1
+    int *dp = (int *)malloc(sizeof(int) * (n + 1));
+    dp[0] = 0, dp[1] = 1, dp[2] = 2;
+
+    //从前向后遍历数组，dp[i] = dp[i-1] + dp[i-2]
+    int i;
+    for(i = 3; i <= n; ++i) {
+        dp[i] = dp[i - 1] + dp[i - 2];
+    }
+    //返回dp[n]
+    return dp[n];
+}
+```
+
+优化空间复杂度：
+```c
+int climbStairs(int n){
+    //若n<=2，返回n
+    if(n <= 2)
+        return n;
+    //初始化dp数组，数组大小为3
+    int *dp = (int *)malloc(sizeof(int) * 3);
+    dp[1] = 1, dp[2] = 2;
+
+    //只记录前面两个台阶的状态
+    int i;
+    for(i = 3; i <= n; ++i) {
+        int sum = dp[1] + dp[2];
+        dp[1] = dp[2];
+        dp[2] = sum;
+    }
+    //返回dp[2]
+    return dp[2];
+}
+```
+
 
 -----------------------
 <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

@@ -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/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/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/0142.环形链表II.md

@@ -294,7 +294,30 @@ var detectCycle = function(head) {
 };
 ```
 
+TypeScript：
+
+```typescript
+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;
+        fastNode = fastNode.next.next;
+        if (slowNode === fastNode) {
+            slowNode = head;
+            while (slowNode !== fastNode) {
+                slowNode = (slowNode as ListNode).next;
+                fastNode = (fastNode as ListNode).next;
+            }
+            return slowNode;
+        }
+    }
+    return null;
+};
+```
+
 Swift:
+
 ```swift
 class Solution {
     func detectCycle(_ head: ListNode?) -> ListNode? {

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/0222.完全二叉树的节点个数.md

@@ -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/0242.有效的字母异位词.md

@@ -214,7 +214,23 @@ var isAnagram = function(s, t) {
 };
 ```
 
+TypeScript：
+
+```typescript
+function isAnagram(s: string, t: string): boolean {
+    if (s.length !== t.length) return false;
+    let helperArr: number[] = new Array(26).fill(0);
+    let pivot: number = 'a'.charCodeAt(0);
+    for (let i = 0, length = s.length; i < length; i++) {
+        helperArr[s.charCodeAt(i) - pivot]++;
+        helperArr[t.charCodeAt(i) - pivot]--;
+    }
+    return helperArr.every(i => i === 0);
+};
+```
+
 Swift：
+
 ```Swift
 func isAnagram(_ s: String, _ t: String) -> Bool {
     if s.count != t.count {

problems/0283.移动零.md

@@ -89,9 +89,33 @@ Python：
         for i in range(slow, len(nums)):
             nums[i] = 0
 ```
+交换前后变量，避免补零
+```python
+    def moveZeroes(self, nums: List[int]) -> None:
+        slow, fast = 0, 0
+        while fast < len(nums):
+            if nums[fast] != 0:
+                nums[slow], nums[fast] = nums[fast], nums[slow]
+                slow += 1 # 保持[0, slow)区间是没有0的
+            fast += 1
+```
 
 Go：
 
+```go
+func moveZeroes(nums []int)  {
+    slow := 0
+    for fast := 0; fast < len(nums); fast ++ {
+        if nums[fast] != 0 {
+            temp := nums[slow]
+            nums[slow] = nums[fast]
+            nums[fast] = temp
+            slow++
+        }
+    }
+}
+```
+
 JavaScript：
 ```javascript
 var moveZeroes = function(nums) {

problems/0452.用最少数量的箭引爆气球.md

@@ -239,5 +239,30 @@ int findMinArrowShots(int** points, int pointsSize, int* pointsColSize){
 }
 ```
 
+### Rust
+```Rust
+use std::cmp;
+impl Solution {
+    pub fn find_min_arrow_shots(mut points: Vec<Vec<i32>>) -> i32 {
+        if points.is_empty() {
+            return 0;
+        }
+        points.sort_by_key(|point| point[0]);
+
+        let size = points.len();
+        let mut count = 1;
+
+         for i in 1..size {
+            if points[i][0] > points[i-1][1] {
+                count += 1;
+            } else {    
+                points[i][1] = cmp::min(points[i][1], points[i-1][1]); 
+            }
+        }
+        
+        return count;
+    }
+}
+```
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0509.斐波那契数.md

@@ -245,6 +245,38 @@ var fib = function(n) {
 };
 ```
 
+### C
+动态规划:
+```c
+int fib(int n){
+    //当n <= 1时，返回n
+    if(n <= 1)
+        return n;
+    //动态开辟一个int数组，大小为n+1
+    int *dp = (int *)malloc(sizeof(int) * (n + 1));
+    //设置0号位为0，1号为为1
+    dp[0] = 0;
+    dp[1] = 1;
+
+    //从前向后遍历数组(i=2; i <= n; ++i),下标为n时的元素为dp[i-1] + dp[i-2]
+    int i;
+    for(i = 2; i <= n; ++i) {
+        dp[i] = dp[i - 1] + dp[i - 2];
+    }
+    return dp[n];
+}
+```
+
+递归实现：
+```c
+int fib(int n){
+    //若n小于等于1，返回n
+    if(n <= 1)
+        return n;
+    //否则返回fib(n-1) + fib(n-2)
+    return fib(n-1) + fib(n-2);
+}
+```
 
 
 -----------------------

problems/0746.使用最小花费爬楼梯.md

@@ -266,5 +266,21 @@ var minCostClimbingStairs = function(cost) {
 };
 ```
 
+### C
+```c
+int minCostClimbingStairs(int* cost, int costSize){
+    //开辟dp数组，大小为costSize
+    int *dp = (int *)malloc(sizeof(int) * costSize);
+    //初始化dp[0] = cost[0], dp[1] = cost[1]
+    dp[0] = cost[0], dp[1] = cost[1];
+
+    int i;
+    for(i = 2; i < costSize; ++i) {
+        dp[i] = (dp[i-1] < dp[i-2] ? dp[i-1] : dp[i-2]) + cost[i];
+    }
+    //选出倒数2层楼梯中较小的
+    return dp[i-1] < dp[i-2] ? dp[i-1] : dp[i-2];
+}
+```
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/面试题02.07.链表相交.md

@@ -239,6 +239,43 @@ var getIntersectionNode = function(headA, headB) {
 };
 ```
 
+TypeScript：
+
+```typescript
+function getIntersectionNode(headA: ListNode | null, headB: ListNode | null): ListNode | null {
+  let sizeA: number = 0,
+    sizeB: number = 0;
+  let curA: ListNode | null = headA,
+    curB: ListNode | null = headB;
+  while (curA) {
+    sizeA++;
+    curA = curA.next;
+  }
+  while (curB) {
+    sizeB++;
+    curB = curB.next;
+  }
+  curA = headA;
+  curB = headB;
+  if (sizeA < sizeB) {
+    [sizeA, sizeB] = [sizeB, sizeA];
+    [curA, curB] = [curB, curA];
+  }
+  let gap = sizeA - sizeB;
+  while (gap-- && curA) {
+    curA = curA.next;
+  }
+  while (curA && curB) {
+    if (curA === curB) {
+      return curA;
+    }
+    curA = curA.next;
+    curB = curB.next;
+  }
+  return null;
+};
+```
+
 C：
 
 ```c