Merge branch 'youngyangyang04:master' into master

Author: SianXiaoCHN

README.md

@@ -1,5 +1,4 @@
 
-
 👉 推荐 [在线阅读](http://programmercarl.com/) (Github在国内访问经常不稳定)         
 👉 推荐 [Gitee同步](https://gitee.com/programmercarl/leetcode-master) 
 

problems/0102.二叉树的层序遍历.md

@@ -1287,23 +1287,23 @@ java代码：
 ```java
 class Solution {
     public List<Integer> largestValues(TreeNode root) {
-		List<Integer> retVal = new ArrayList<Integer>();
-		Queue<TreeNode> tmpQueue = new LinkedList<TreeNode>();
-		if (root != null) tmpQueue.add(root);
-		
-		while (tmpQueue.size() != 0){
-			int size = tmpQueue.size();
-			List<Integer> lvlVals = new ArrayList<Integer>();
-			for (int index = 0; index < size; index++){
-				TreeNode node = tmpQueue.poll();
-				lvlVals.add(node.val);
-				if (node.left != null) tmpQueue.add(node.left);
-				if (node.right != null) tmpQueue.add(node.right);
-			}
-			retVal.add(Collections.max(lvlVals));
-		}
-        
-        return retVal;
+        if(root == null){
+            return Collections.emptyList();
+        }
+        List<Integer> result = new ArrayList();
+        Queue<TreeNode> queue = new LinkedList();
+        queue.offer(root);
+        while(!queue.isEmpty()){
+            int max = Integer.MIN_VALUE;
+            for(int i = queue.size(); i > 0; i--){
+               TreeNode node = queue.poll();
+               max = Math.max(max, node.val);
+               if(node.left != null) queue.offer(node.left);
+               if(node.right != null) queue.offer(node.right);
+            }
+            result.add(max);
+        } 
+        return result;
     }
 }
 ```

problems/0309.最佳买卖股票时机含冷冻期.md

@@ -205,6 +205,29 @@ class Solution {
     }
 }
 ```
+```java
+//另一种解题思路
+class Solution {
+    public int maxProfit(int[] prices) {
+        int[][] dp = new int[prices.length + 1][2];
+        dp[1][0] = -prices[0];
+
+        for (int i = 2; i <= prices.length; i++) {
+            /*
+            dp[i][0] 第i天未持有股票收益;
+            dp[i][1] 第i天持有股票收益;
+            情况一：第i天是冷静期，不能以dp[i-1][1]购买股票,所以以dp[i - 2][1]买股票，没问题
+            情况二：第i天不是冷静期，理论上应该以dp[i-1][1]购买股票，但是第i天不是冷静期说明，第i-1天没有卖出股票，
+                则dp[i-1][1]=dp[i-2][1],所以可以用dp[i-2][1]买股票，没问题
+             */
+            dp[i][0] = Math.max(dp[i - 1][0], dp[i - 2][1] - prices[i - 1]);
+            dp[i][1] = Math.max(dp[i - 1][1], dp[i - 1][0] + prices[i - 1]);
+        }
+
+        return dp[prices.length][1];
+    }
+}
+```
 
 Python：
 

problems/0416.分割等和子集.md

@@ -251,14 +251,14 @@ Python：
 ```python
 class Solution:
     def canPartition(self, nums: List[int]) -> bool:
-        taraget = sum(nums)
-        if taraget % 2 == 1: return False
-        taraget //= 2
+        target = sum(nums)
+        if target % 2 == 1: return False
+        target //= 2
         dp = [0] * 10001
         for i in range(len(nums)):
-            for j in range(taraget, nums[i] - 1, -1):
+            for j in range(target, nums[i] - 1, -1):
                 dp[j] = max(dp[j], dp[j - nums[i]] + nums[i])
-        return taraget == dp[taraget]
+        return target == dp[target]
 ```
 Go：
 ```go