Merge branch 'youngyangyang04:master' into master

problems/0108.将有序数组转换为二叉搜索树.md

@@ -209,6 +209,8 @@ public:
 
 
 Java：
+
+递归: 左闭右开 [left,right)
 ```Java
 class Solution {
     public TreeNode sortedArrayToBST(int[] nums) {
@@ -232,6 +234,75 @@ class Solution {
 
 ```
 
+递归: 左闭右闭 [left,right]
+```java
+class Solution {
+	public TreeNode sortedArrayToBST(int[] nums) {
+		TreeNode root = traversal(nums, 0, nums.length - 1);
+		return root;
+	}
+
+	// 左闭右闭区间[left, right)
+	private TreeNode traversal(int[] nums, int left, int right) {
+		if (left > right) return null;
+
+		int mid = left + ((right - left) >> 1);
+		TreeNode root = new TreeNode(nums[mid]);
+		root.left = traversal(nums, left, mid - 1);
+		root.right = traversal(nums, mid + 1, right);
+		return root;
+	}
+}
+```
+迭代: 左闭右闭 [left,right]
+```java
+class Solution {
+	public TreeNode sortedArrayToBST(int[] nums) {
+		if (nums.length == 0) return null;
+
+		//根节点初始化
+		TreeNode root = new TreeNode(-1);
+		Queue<TreeNode> nodeQueue = new LinkedList<>();
+		Queue<Integer> leftQueue = new LinkedList<>();
+		Queue<Integer> rightQueue = new LinkedList<>();
+
+		// 根节点入队列
+		nodeQueue.offer(root);
+		// 0为左区间下表初始位置
+		leftQueue.offer(0);
+		// nums.size() - 1为右区间下表初始位置
+		rightQueue.offer(nums.length - 1);
+
+		while (!nodeQueue.isEmpty()) {
+			TreeNode currNode = nodeQueue.poll();
+			int left = leftQueue.poll();
+			int right = rightQueue.poll();
+			int mid = left + ((right - left) >> 1);
+
+			// 将mid对应的元素给中间节点
+			currNode.val = nums[mid];
+
+			// 处理左区间
+			if (left <= mid - 1) {
+				currNode.left = new TreeNode(-1);
+				nodeQueue.offer(currNode.left);
+				leftQueue.offer(left);
+				rightQueue.offer(mid - 1);
+			}
+
+			// 处理右区间
+			if (right >= mid + 1) {
+				currNode.right = new TreeNode(-1);
+				nodeQueue.offer(currNode.right);
+				leftQueue.offer(mid + 1);
+				rightQueue.offer(right);
+			}
+		}
+		return root;
+	}
+}
+```
+
 Python：
 ```python3
 # Definition for a binary tree node.

problems/0115.不同的子序列.md

@@ -186,6 +186,40 @@ class Solution:
         return dp[-1][-1]
 ```
 
+Python3:
+```python
+class SolutionDP2:
+    """
+    既然dp[i]只用到dp[i - 1]的状态，
+    我们可以通过缓存dp[i - 1]的状态来对dp进行压缩，
+    减少空间复杂度。
+    （原理等同同于滚动数组）
+    """
+    
+    def numDistinct(self, s: str, t: str) -> int:
+        n1, n2 = len(s), len(t)
+        if n1 < n2:
+            return 0
+
+        dp = [0 for _ in range(n2 + 1)]
+        dp[0] = 1
+
+        for i in range(1, n1 + 1):
+            # 必须深拷贝
+            # 不然prev[i]和dp[i]是同一个地址的引用
+            prev = dp.copy()
+            # 剪枝，保证s的长度大于等于t
+            # 因为对于任意i，i > n1, dp[i] = 0
+            # 没必要跟新状态。 
+            end = i if i < n2 else n2
+            for j in range(1, end + 1):
+                if s[i - 1] == t[j - 1]:
+                    dp[j] = prev[j - 1] + prev[j]
+                else:
+                    dp[j] = prev[j]
+        return dp[-1]
+```
+
 Go：
 
 

problems/周总结/20201003二叉树周末总结.md

@@ -40,9 +40,8 @@ public:
         return isSame;
 
     }
-    bool isSymmetric(TreeNode* root) {
+    bool isSymmetric(TreeNode* p, TreeNode* q) {
-        if (root == NULL) return true;
+        return compare(p, q);
-        return compare(root->left, root->right);
     }
 };
 ```