Merge branch 'youngyangyang04:master' into master

problems/0019.删除链表的倒数第N个节点.md

@@ -129,10 +129,10 @@ class Solution:
         head_dummy.next = head
 
         slow, fast = head_dummy, head_dummy
-        while(n!=0): #fast先往前走n步
+        while(n>=0): #fast先往前走n+1步
             fast = fast.next
             n -= 1
-        while(fast.next!=None):
+        while(fast!=None):
             slow = slow.next
             fast = fast.next
         #fast 走到结尾后，slow的下一个节点为倒数第N个节点

problems/0059.螺旋矩阵II.md

@@ -117,6 +117,9 @@ public:
 };
 ```
 
+* 时间复杂度 O(n^2): 模拟遍历二维矩阵的时间
+* 空间复杂度 O(1)
+
 ## 类似题目
 
 * 54.螺旋矩阵

problems/0455.分发饼干.md

@@ -219,19 +219,17 @@ func findContentChildren(g []int, s []int) int {
 
 ### Rust
 ```rust
-pub fn find_content_children(children: Vec<i32>, cookie: Vec<i32>) -> i32 {
+pub fn find_content_children(mut children: Vec<i32>, mut cookie: Vec<i32>) -> i32 {
-    let mut children = children;
-    let mut cookies = cookie;
     children.sort();
     cookies.sort();
 
-    let (mut child, mut cookie) = (0usize, 0usize);
+    let (mut child, mut cookie) = (0, 0);
     while child < children.len() && cookie < cookies.len() {
         // 优先选择最小饼干喂饱孩子
         if children[child] <= cookies[cookie] {
             child += 1;
         }
-        cookie += 1
+        cookie += 1;
     }
     child as i32
 }

problems/0704.二分查找.md

@@ -86,6 +86,9 @@ public:
 };
 
 ```
+* 时间复杂度：O(log n)
+* 空间复杂度：O(1)
+
 
 ### 二分法第二种写法
 
@@ -124,6 +127,9 @@ public:
     }
 };
 ```
+* 时间复杂度：O(log n)
+* 空间复杂度：O(1)
+
 
 ## 总结
 

problems/前序/ACM模式如何构建二叉树.md

@@ -305,11 +305,12 @@ def construct_binary_tree(nums: []) -> TreeNode:
         Tree.append(node)
         if i == 0:
             root = node
+    # 直接判断2*i+2<len(Tree)会漏掉2*i+1=len(Tree)-1的情况
     for i in range(len(Tree)):
-        node = Tree[i]
+        if Tree[i] and 2 * i + 1 < len(Tree):
-        if node and (2 * i + 2) < len(Tree):
+            Tree[i].left = Tree[2 * i + 1]
-            node.left = Tree[i * 2 + 1]
+            if 2 * i + 2 < len(Tree):
-            node.right = Tree[i * 2 + 2]
+                Tree[i].right = Tree[2 * i + 2]
     return root