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

problems/0027.移除元素.md

@@ -185,7 +185,7 @@ func removeElement(nums []int, val int) int {
 ```
 
 JavaScript:
-```
+```javascript
 //时间复杂度O(n)
 //空间复杂度O(1)
 var removeElement = (nums, val) => {

problems/0077.组合.md

@@ -368,6 +368,34 @@ class Solution {
 }
 ```
 
+Python2:
+```python
+class Solution(object):
+    def combine(self, n, k):
+        """
+        :type n: int
+        :type k: int
+        :rtype: List[List[int]]
+        """
+        result = []
+        path = []
+        def backtracking(n, k, startidx):
+            if len(path) == k:
+                result.append(path[:])
+                return
+
+            # 剪枝， 最后k - len(path)个节点直接构造结果，无需递归
+            last_startidx = n - (k - len(path)) + 1
+            result.append(path + [idx for idx in range(last_startidx, n + 1)])
+
+            for x in range(startidx, last_startidx):
+                path.append(x)
+                backtracking(n, k, x + 1)  # 递归
+                path.pop()  # 回溯
+
+        backtracking(n, k, 1)
+        return result
+```
 
 ## Python 
 ```python

problems/0078.子集.md

@@ -263,6 +263,63 @@ var subsets = function(nums) {
 };
 ```
 
+C:
+```c
+int* path;
+int pathTop;
+int** ans;
+int ansTop;
+//记录二维数组中每个一维数组的长度
+int* length;
+//将当前path数组复制到ans中
+void copy() {
+    int* tempPath = (int*)malloc(sizeof(int) * pathTop);
+    int i;
+    for(i = 0; i < pathTop; i++) {
+        tempPath[i] = path[i];
+    }
+    ans = (int**)realloc(ans, sizeof(int*) * (ansTop+1));
+    length[ansTop] = pathTop;
+    ans[ansTop++] = tempPath;
+}
+
+void backTracking(int* nums, int numsSize, int startIndex) {
+    //收集子集，要放在终止添加的上面，否则会漏掉自己
+    copy();
+    //若startIndex大于数组大小，返回
+    if(startIndex >= numsSize) {
+        return;
+    }
+    int j;
+    for(j = startIndex; j < numsSize; j++) {
+        //将当前下标数字放入path中
+        path[pathTop++] = nums[j];
+        backTracking(nums, numsSize, j+1);
+        //回溯
+        pathTop--;
+    }
+}
+
+int** subsets(int* nums, int numsSize, int* returnSize, int** returnColumnSizes){
+    //初始化辅助变量
+    path = (int*)malloc(sizeof(int) * numsSize);
+    ans = (int**)malloc(0);
+    length = (int*)malloc(sizeof(int) * 1500);
+    ansTop = pathTop = 0;
+    //进入回溯
+    backTracking(nums, numsSize, 0);
+    //设置二维数组中元素个数
+    *returnSize = ansTop;
+    //设置二维数组中每个一维数组的长度
+    *returnColumnSizes = (int*)malloc(sizeof(int) * ansTop);
+    int i;
+    for(i = 0; i < ansTop; i++) {
+        (*returnColumnSizes)[i] = length[i];
+    }
+    return ans;
+}
+```
+
 
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)

problems/0093.复原IP地址.md

@@ -476,6 +476,82 @@ func isNormalIp(s string,startIndex,end int)bool{
 
 ```
 
+C:
+```c
+//记录结果
+char** result;
+int resultTop;
+//记录应该加入'.'的位置
+int segments[3];
+int isValid(char* s, int start, int end) {
+    if(start > end)
+        return 0;
+    if (s[start] == '0' && start != end) { // 0开头的数字不合法
+                return false;
+    }
+    int num = 0;
+    for (int i = start; i <= end; i++) {
+        if (s[i] > '9' || s[i] < '0') { // 遇到非数字字符不合法
+            return false;
+        }
+        num = num * 10 + (s[i] - '0');
+        if (num > 255) { // 如果大于255了不合法
+            return false;
+        }
+    }
+    return true;
+}
+
+//startIndex为起始搜索位置，pointNum为'.'对象
+void backTracking(char* s, int startIndex, int pointNum) {
+    //若'.'数量为3，分隔结束
+    if(pointNum == 3) {
+        //若最后一段字符串符合要求，将当前的字符串放入result种
+        if(isValid(s, startIndex, strlen(s) - 1)) {
+            char* tempString = (char*)malloc(sizeof(char) * strlen(s) + 4);
+            int j;
+            //记录添加字符时tempString的下标
+            int count = 0;
+            //记录添加字符时'.'的使用数量
+            int count1 = 0;
+            for(j = 0; j < strlen(s); j++) {
+                tempString[count++] = s[j];
+                //若'.'的使用数量小于3且当前下标等于'.'下标，添加'.'到数组
+                if(count1 < 3 && j == segments[count1]) {
+                    tempString[count++] = '.';
+                    count1++;
+                }
+            }
+            tempString[count] = 0;
+            //扩容result数组
+            result = (char**)realloc(result, sizeof(char*) * (resultTop + 1));
+            result[resultTop++] = tempString;
+        }
+        return ;
+    }
+
+    int i;
+    for(i = startIndex; i < strlen(s); i++) {
+        if(isValid(s, startIndex, i)) {
+            //记录应该添加'.'的位置
+            segments[pointNum] = i;
+            backTracking(s, i + 1, pointNum + 1);
+        }
+        else {
+            break;
+        }
+    }
+}
+
+char ** restoreIpAddresses(char * s, int* returnSize){
+    result = (char**)malloc(0);
+    resultTop = 0;
+    backTracking(s, 0, 0);
+    *returnSize = resultTop;
+    return result;
+}
+```
+
 
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)

problems/0101.对称二叉树.md

@@ -185,7 +185,8 @@ public:
         queue<TreeNode*> que;
         que.push(root->left);   // 将左子树头结点加入队列
         que.push(root->right);  // 将右子树头结点加入队列
-        while (!que.empty()) {  // 接下来就要判断这这两个树是否相互翻转
+
+        while (!que.empty()) {  // 接下来就要判断这两个树是否相互翻转
             TreeNode* leftNode = que.front(); que.pop();
             TreeNode* rightNode = que.front(); que.pop();
             if (!leftNode && !rightNode) {  // 左节点为空、右节点为空，此时说明是对称的

problems/0188.买卖股票的最佳时机IV.md

@@ -280,6 +280,7 @@ Go：
 Javascript：
 
 ```javascript
+// 方法一：动态规划
 const maxProfit = (k,prices) => {
     if (prices == null || prices.length < 2 || k == 0) {
         return 0;
@@ -300,6 +301,30 @@ const maxProfit = (k,prices) => {
 
     return dp[prices.length - 1][2 * k];
 };
+
+// 方法二：动态规划+空间优化
+var maxProfit = function(k, prices) {
+    let n = prices.length;
+    let dp = new Array(2*k+1).fill(0);
+    // dp 买入状态初始化
+    for (let i = 1; i <= 2*k; i += 2) {
+        dp[i] = - prices[0];
+    }
+
+    for (let i = 1; i < n; i++) {
+        for (let j = 1; j < 2*k+1; j++) {
+            // j 为奇数：买入状态
+            if (j % 2) {
+                dp[j] = Math.max(dp[j], dp[j-1] - prices[i]);
+            } else {
+                // j为偶数：卖出状态
+                dp[j] = Math.max(dp[j], dp[j-1] + prices[i]);
+            }
+        }
+    }
+
+    return dp[2*k];
+};
 ```
 
 -----------------------

problems/0202.快乐数.md

@@ -159,26 +159,28 @@ func getSum(n int) int {
 javaScript:
 
 ```js
-function getN(n) {
-    if (n == 1 || n == 0) return n;
-    let res = 0;
-    while (n) {
-        res += (n % 10) * (n % 10);
-        n = parseInt(n / 10);
+var isHappy = function (n) {
+    let m = new Map()
+
+    const getSum = (num) => {
+        let sum = 0
+        while (n) {
+            sum += (n % 10) ** 2
+            n = Math.floor(n / 10)
+        }
+        return sum
     }
-    return res;
-}
 
-var isHappy = function(n) {
-    const sumSet = new Set();
-    while (n != 1 && !sumSet.has(n)) {
-        sumSet.add(n);
-        n = getN(n);
+    while (true) {
+        // n出现过，证明已陷入无限循环
+        if (m.has(n)) return false
+        if (n === 1) return true
+        m.set(n, 1)
+        n = getSum(n)
     }
-    return n == 1;
-};
+}
 
-// 使用环形链表的思想 说明出现闭环 退出循环
+// 方法二：使用环形链表的思想 说明出现闭环 退出循环
 var isHappy = function(n) {
     if (getN(n) == 1) return true;
     let a = getN(n), b = getN(getN(n));

problems/0235.二叉搜索树的最近公共祖先.md

@@ -268,17 +268,30 @@ class Solution {
 递归法：
 ```python
 class Solution:
+    """二叉搜索树的最近公共祖先 递归法"""
+
     def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
-        if not root: return root  //中
-        if root.val >p.val and root.val > q.val:
-            return self.lowestCommonAncestor(root.left,p,q)  //左
-        elif root.val < p.val and root.val < q.val:
-            return self.lowestCommonAncestor(root.right,p,q)  //右
-        else: return root
+        if root.val > p.val and root.val > q.val:
+            return self.lowestCommonAncestor(root.left, p, q)
+        if root.val < p.val and root.val < q.val:
+            return self.lowestCommonAncestor(root.right, p, q)
+        return root
 ```
 
 迭代法：
+```python
+class Solution:
+    """二叉搜索树的最近公共祖先 迭代法"""
 
+    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
+        while True:
+            if root.val > p.val and root.val > q.val:
+                root = root.left
+            elif root.val < p.val and root.val < q.val:
+                root = root.right
+            else:
+                return root
+```
 
 ## Go 
 

problems/0236.二叉树的最近公共祖先.md

@@ -264,16 +264,21 @@ class Solution {
 ## Python 
 
 ```python
-//递归
 class Solution:
+    """二叉树的最近公共祖先 递归法"""
+
     def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
-        if not root or root == p or root == q: return root  //找到了节点p或者q，或者遇到空节点
-        left = self.lowestCommonAncestor(root.left,p,q)  //左
-        right = self.lowestCommonAncestor(root.right,p,q)  //右
-        if left and right: return root  //中: left和right不为空，root就是最近公共节点
-        elif left and not right: return left  //目标节点是通过left返回的
-        elif not left and right: return right  //目标节点是通过right返回的
-        else: return None  //没找到
+        if not root or root == p or root == q:
+            return root
+
+        left = self.lowestCommonAncestor(root.left, p, q)
+        right = self.lowestCommonAncestor(root.right, p, q)
+
+        if left and right:
+            return root
+        if left:
+            return left
+        return right
 ```
 
 ## Go 

problems/0455.分发饼干.md

@@ -217,6 +217,29 @@ var findContentChildren = function(g, s) {
 
 ```
 
+C:
+```c
+int cmp(int* a, int* b) {
+    return *a - *b;
+}
+
+int findContentChildren(int* g, int gSize, int* s, int sSize){
+    if(sSize == 0)
+        return 0;
+
+    //将两个数组排序为升序
+    qsort(g, gSize, sizeof(int), cmp);
+    qsort(s, sSize, sizeof(int), cmp);
+
+    int numFedChildren = 0;
+    int i = 0;
+    for(i = 0; i < sSize; ++i) {
+        if(numFedChildren < gSize && g[numFedChildren] <= s[i])
+            numFedChildren++;
+    }
+    return numFedChildren;
+}
+```
 
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)