Merge branch 'master' into master

Author: youngyangyang04

problems/0077.组合优化.md

@@ -242,8 +242,59 @@ var combine = function(n, k) {
 };
 ```
 
+C:
+```c
+int* path;
+int pathTop;
+int** ans;
+int ansTop;
+
+void backtracking(int n, int k,int startIndex) {
+    //当path中元素个数为k个时，我们需要将path数组放入ans二维数组中
+    if(pathTop == k) {
+        //path数组为我们动态申请，若直接将其地址放入二维数组，path数组中的值会随着我们回溯而逐渐变化
+        //因此创建新的数组存储path中的值
+        int* temp = (int*)malloc(sizeof(int) * k);
+        int i;
+        for(i = 0; i < k; i++) {
+            temp[i] = path[i];
+        }
+        ans[ansTop++] = temp;
+        return ;
+    }
+
+    int j;
+    for(j = startIndex; j <= n- (k - pathTop) + 1;j++) {
+        //将当前结点放入path数组
+        path[pathTop++] = j;
+        //进行递归
+        backtracking(n, k, j + 1);
+        //进行回溯，将数组最上层结点弹出
+        pathTop--;
+    }
+}
 
+int** combine(int n, int k, int* returnSize, int** returnColumnSizes){
+    //path数组存储符合条件的结果
+    path = (int*)malloc(sizeof(int) * k);
+    //ans二维数组存储符合条件的结果数组的集合。（数组足够大，避免极端情况）
+    ans = (int**)malloc(sizeof(int*) * 10000);
+    pathTop = ansTop = 0;
 
+    //回溯算法
+    backtracking(n, k, 1);
+    //最后的返回大小为ans数组大小
+    *returnSize = ansTop;
+    //returnColumnSizes数组存储ans二维数组对应下标中一维数组的长度（都为k）
+    *returnColumnSizes = (int*)malloc(sizeof(int) *(*returnSize));
+    int i;
+    for(i = 0; i < *returnSize; i++) {
+        (*returnColumnSizes)[i] = k;
+    }
+    //返回ans二维数组
+    return ans;
+}
+```
 
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)

problems/0102.二叉树的层序遍历.md

@@ -87,26 +87,31 @@ public:
 
 python代码：
 
-```python
+```python3
 class Solution:
+    """二叉树层序遍历迭代解法"""
+
     def levelOrder(self, root: TreeNode) -> List[List[int]]:
+        results = []
         if not root:
-            return []
-
-        queue = [root]
-        out_list = []
-
-        while queue:
-            length = len(queue)  
-            in_list = []
-            for _ in range(length):
-                curnode = queue.pop(0)  # （默认移除列表最后一个元素）这里需要移除队列最头上的那个
-                in_list.append(curnode.val)
-                if curnode.left: queue.append(curnode.left)
-                if curnode.right: queue.append(curnode.right)
-            out_list.append(in_list)
+            return results
+        
+        from collections import deque
+        que = deque([root])
+        
+        while que:
+            size = len(que)
+            result = []
+            for _ in range(size):
+                cur = que.popleft()
+                result.append(cur.val)
+                if cur.left:
+                    que.append(cur.left)
+                if cur.right:
+                    que.append(cur.right)
+            results.append(result)
 
-        return out_list
+        return results
 ```
 
 java: 
@@ -274,29 +279,29 @@ python代码：
 
 ```python
 class Solution:
+    """二叉树层序遍历II迭代解法"""
+
     def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
+        results = []
         if not root:
-            return []
-        quene = [root]
-        out_list = []
+            return results
 
-        while quene:
-            in_list = []
-            for _ in range(len(quene)):
-                node = quene.pop(0)
-                in_list.append(node.val)
-                if node.left:
-                    quene.append(node.left)
-                if node.right:
-                    quene.append(node.right)
- 
-            out_list.append(in_list)
-    
-        out_list.reverse()
-        return out_list
-    
-# 执行用时：36 ms, 在所有 Python3 提交中击败了92.00%的用户
-# 内存消耗：15.2 MB, 在所有 Python3 提交中击败了63.76%的用户
+        from collections import deque
+        que = deque([root])
+        
+        while que:
+            result = []
+            for _ in range(len(que)):
+                cur = que.popleft()
+                result.append(cur.val)
+                if cur.left:
+                    que.append(cur.left)
+                if cur.right:
+                    que.append(cur.right)
+            results.append(result)
+
+        results.reverse()
+        return results
 ```
 
 Java：
@@ -628,32 +633,29 @@ python代码：
 
 ```python
 class Solution:
+    """二叉树层平均值迭代解法"""
+
     def averageOfLevels(self, root: TreeNode) -> List[float]:
+        results = []
         if not root:
-            return []
+            return results
         
-        quene = deque([root])
-        out_list = []
-
-        while quene:
-            in_list = []
-
-            for _ in range(len(quene)):
-                node = quene.popleft()
-                in_list.append(node.val)
-                if node.left:
-                    quene.append(node.left)
-                if node.right:
-                    quene.append(node.right)
-
-            out_list.append(in_list)
-
-        out_list = map(lambda x: sum(x) / len(x), out_list)
-    
-        return out_list
+        from collections import deque
+        que = deque([root])
         
-# 执行用时：56 ms, 在所有 Python3 提交中击败了81.48%的用户
-# 内存消耗：17 MB, 在所有 Python3 提交中击败了89.68%的用户
+        while que:
+            size = len(que)
+            sum_ = 0
+            for _ in range(size):
+                cur = que.popleft()
+                sum_ += cur.val
+                if cur.left:
+                    que.append(cur.left)
+                if cur.right:
+                    que.append(cur.right)
+            results.append(sum_ / size)
+
+        return results
 ```
 
 java:
@@ -823,52 +825,28 @@ public:
 python代码：
 
 ```python
-
 class Solution:
+    """N叉树的层序遍历迭代法"""
+
     def levelOrder(self, root: 'Node') -> List[List[int]]:
+        results = []
         if not root:
-            return []
+            return results
 
-        quene = deque([root])
-        out_list = []
-
-        while quene:
-            in_list = []
-
-            for _ in range(len(quene)):
-                node = quene.popleft()
-                in_list.append(node.val)
-                if node.children:
-                    # 这个地方要用extend而不是append，我们看下面的例子：
-                    # In [18]: alist=[]
-                    # In [19]: alist.append([1,2,3])
-                    # In [20]: alist
-                    # Out[20]: [[1, 2, 3]]
-                    # In [21]: alist.extend([4,5,6])
-                    # In [22]: alist
-                    # Out[22]: [[1, 2, 3], 4, 5, 6]
-                    # 可以看到extend对要添加的list进行了一个解包操作
-                    # print(root.children)，可以得到children是一个包含
-                    # 孩子节点地址的list，我们使用for遍历quene的时候，
-                    # 希望quene是一个单层list，所以要用extend
-                    # 使用extend的情况，如果print(quene),结果是
-                    # deque([<__main__.Node object at 0x7f60763ae0a0>])
-				   # deque([<__main__.Node object at 0x7f607636e6d0>, <__main__.Node object at 0x7f607636e130>, <__main__.Node object at 0x7f607636e310>])
-				  # deque([<__main__.Node object at 0x7f607636e880>, <__main__.Node object at 0x7f607636ef10>])
-				  # 可以看到是单层list
-                    # 如果使用append，print(quene)的结果是
-                    # deque([<__main__.Node object at 0x7f18907530a0>])
-				  # deque([[<__main__.Node object at 0x7f18907136d0>, <__main__.Node object at 0x7f1890713130>, <__main__.Node object at 0x7f1890713310>]])
-				  # 可以看到是两层list，这样for的遍历就会报错
-                    
-                    quene.extend(node.children)
-                
-            out_list.append(in_list)
+        from collections import deque
+        que = deque([root])
 
-        return out_list
-    
-# 执行用时：60 ms, 在所有 Python3 提交中击败了76.99%的用户
-# 内存消耗：16.5 MB, 在所有 Python3 提交中击败了89.19%的用户
+        while que:
+            result = []
+            for _ in range(len(que)):
+                cur = que.popleft()
+                result.append(cur.val)
+                # cur.children 是 Node 对象组成的列表，也可能为 None
+                if cur.children:
+                    que.extend(cur.children)
+            results.append(result)
+
+        return results
 ```
 
 java:

problems/0383.赎金信.md

@@ -266,6 +266,7 @@ var canConstruct = function(ransomNote, magazine) {
 };
 ```
 
+
 PHP:
 ```php
 class Solution {
@@ -289,6 +290,28 @@ class Solution {
         }
         return true;
     }
+```
+
+Swift：
+```swift
+func canConstruct(_ ransomNote: String, _ magazine: String) -> Bool {
+    var record = Array(repeating: 0, count: 26);
+    let aUnicodeScalarValue = "a".unicodeScalars.first!.value
+    for unicodeScalar in magazine.unicodeScalars {
+        // 通过record 记录 magazine 里各个字符出现的次数
+        let idx: Int = Int(unicodeScalar.value - aUnicodeScalarValue)
+        record[idx] += 1
+    }
+    for unicodeScalar in ransomNote.unicodeScalars {
+        // 遍历 ransomNote,在record里对应的字符个数做 -- 操作
+        let idx: Int = Int(unicodeScalar.value - aUnicodeScalarValue)
+        record[idx] -= 1
+        // 如果小于零说明在magazine没有
+        if record[idx] < 0 {
+            return false
+        }
+    }
+    return true
 }
 ```
 

problems/0454.四数相加II.md

@@ -220,6 +220,7 @@ var fourSumCount = function(nums1, nums2, nums3, nums4) {
 };
 ```
 
+
 PHP:
 ```php
 class Solution {
@@ -253,6 +254,35 @@ 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
+    var map = [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 i in 0 ..< nums1.count {
+        for j in 0 ..< nums2.count {
+            let sum1 = nums1[i] + nums2[j]
+            map[sum1] = (map[sum1] ?? 0) + 1
+        }
+    }
+    // 统计a+b+c+d = 0 出现的次数
+    var res = 0
+    // 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。
+    for i in 0 ..< nums3.count {
+        for j in 0 ..< nums4.count {
+            let sum2 = nums3[i] + nums4[j]
+            let other = 0 - sum2
+            if map.keys.contains(other) {
+                res += map[other]!
+            }
+        }
+    }
+    return res
+}
+```
+
+
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
 * B站视频：[代码随想录](https://space.bilibili.com/525438321)

problems/0455.分发饼干.md

@@ -197,11 +197,10 @@ func findContentChildren(g []int, s []int) int {
 
   return child
 }
-
+```
 
 Javascript:
-​```Javascript
-
+```
 var findContentChildren = function(g, s) {
     g = g.sort((a, b) => a - b)
     s = s.sort((a, b) => a - b)