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

problems/0028.实现strStr.md

@@ -9,7 +9,7 @@
 
 # 28. 实现 strStr()
 
-[力扣题目链接](https://leetcode.cn/problems/implement-strstr/)
+[力扣题目链接](https://leetcode.cn/problems/find-the-index-of-the-first-occurrence-in-a-string/)
 
 实现 strStr() 函数。
 

problems/0031.下一个排列.md

@@ -160,73 +160,34 @@ class Solution {
 ```
 
 ## Python
->直接使用sorted()不符合题意
+>直接使用sorted()会开辟新的空间并返回一个新的list，故补充一个原地反转函数
 ```python
 class Solution:
-    def nextPermutation(self, nums: List[int]) -> None:
-        """
-        Do not return anything, modify nums in-place instead.
-        """
-        for i in range(len(nums)-1, -1, -1):
-            for j in range(len(nums)-1, i, -1):
-                if nums[j] > nums[i]:
-                    nums[j], nums[i] = nums[i], nums[j]
-                    nums[i+1:len(nums)] = sorted(nums[i+1:len(nums)])
-                    return
-        nums.sort()
-```
->另一种思路
-```python
-class Solution:
-    '''
-    抛砖引玉：因题目要求“必须原地修改，只允许使用额外常数空间”，python内置sorted函数以及数组切片+sort()无法使用。  
-    故选择另一种算法暂且提供一种python思路
-    '''
     def nextPermutation(self, nums: List[int]) -> None:
         """
         Do not return anything, modify nums in-place instead.
         """
         length = len(nums)
-        for i in range(length-1, 0, -1):
+        for i in range(length - 1, -1, -1):
-            if nums[i-1] < nums[i]:
+            for j in range(length - 1, i, -1):
-                for j in range(length-1, 0, -1):
+                if nums[j] > nums[i]:
-                    if nums[j] > nums[i-1]:
+                    nums[j], nums[i] = nums[i], nums[j]
-                        nums[i-1], nums[j] = nums[j], nums[i-1]
+                    self.reverse(nums, i + 1, length - 1)
-                        break
+                    return  
-                self.reverse(nums, i, length-1)
+        self.reverse(nums, 0, length - 1)
-                break
-            if n == 1:
-                # 若正常结束循环，则对原数组直接翻转
-                self.reverse(nums, 0, length-1)
     
-    def reverse(self, nums: List[int], low: int, high: int) -> None:
+    def reverse(self, nums: List[int], left: int, right: int) -> None:
-        while low < high:
+        while left < right:
-            nums[low], nums[high] = nums[high], nums[low]
+            nums[left], nums[right] = nums[right], nums[left]
-            low += 1
+            left += 1
-            high -= 1
+            right -= 1
-```
->上一版本简化版
-```python
-class Solution(object):
-    def nextPermutation(self, nums: List[int]) -> None:
-        n = len(nums)
-        i = n-2
-        while i >= 0 and nums[i] >= nums[i+1]:
-            i -= 1
             
-        if i > -1: // i==-1,不存在下一个更大的排列
+"""
-            j = n-1
+265 / 265 个通过测试用例
-            while j >= 0 and nums[j] <= nums[i]:
+状态：通过
-                j -= 1
+执行用时: 36 ms
-            nums[i], nums[j] = nums[j], nums[i]
+内存消耗: 14.9 MB
-    
+"""
-        start, end = i+1, n-1
-        while start < end:
-            nums[start], nums[end] = nums[end], nums[start]
-            start += 1
-            end -= 1
-        
-        return nums
 ```
 
 ## Go

problems/0040.组合总和II.md

@@ -258,40 +258,45 @@ public:
 **使用标记数组**
 ```Java
 class Solution {
-    List<List<Integer>> lists = new ArrayList<>();
+  LinkedList<Integer> path = new LinkedList<>();
-    Deque<Integer> deque = new LinkedList<>();
+  List<List<Integer>> ans = new ArrayList<>();
+  boolean[] used;
   int sum = 0;
 
   public List<List<Integer>> combinationSum2(int[] candidates, int target) {
-        //为了将重复的数字都放到一起，所以先进行排序
+    used = new boolean[candidates.length];
+    // 加标志数组，用来辅助判断同层节点是否已经遍历
+    Arrays.fill(used, false);
+    // 为了将重复的数字都放到一起，所以先进行排序
     Arrays.sort(candidates);
-        //加标志数组，用来辅助判断同层节点是否已经遍历
+    backTracking(candidates, target, 0);
-        boolean[] flag = new boolean[candidates.length];
+    return ans;
-        backTracking(candidates, target, 0, flag);
-        return lists;
   }
 
-    public void backTracking(int[] arr, int target, int index, boolean[] flag) {
+  private void backTracking(int[] candidates, int target, int startIndex) {
     if (sum == target) {
-            lists.add(new ArrayList(deque));
+      ans.add(new ArrayList(path));
-            return;
     }
-        for (int i = index; i < arr.length && arr[i] + sum <= target; i++) {
+    for (int i = startIndex; i < candidates.length; i++) {
-            //出现重复节点，同层的第一个节点已经被访问过，所以直接跳过
+      if (sum + candidates[i] > target) {
-            if (i > 0 && arr[i] == arr[i - 1] && !flag[i - 1]) {
+        break;
+      }
+      // 出现重复节点，同层的第一个节点已经被访问过，所以直接跳过
+      if (i > 0 && candidates[i] == candidates[i - 1] && !used[i - 1]) {
         continue;
       }
-            flag[i] = true;
+      used[i] = true;
-            sum += arr[i];
+      sum += candidates[i];
-            deque.push(arr[i]);
+      path.add(candidates[i]);
-            //每个节点仅能选择一次，所以从下一位开始
+      // 每个节点仅能选择一次，所以从下一位开始
-            backTracking(arr, target, i + 1, flag);
+      backTracking(candidates, target, i + 1);
-            int temp = deque.pop();
+      used[i] = false;
-            flag[i] = false;
+      sum -= candidates[i];
-            sum -= temp;
+      path.removeLast();
     }
   }
 }
+
 ```
 **不使用标记数组**
 ```Java

problems/0046.全排列.md

@@ -331,7 +331,7 @@ var permute = function(nums) {
 
 ```
 
-## TypeScript
+### TypeScript
 
 ```typescript
 function permute(nums: number[]): number[][] {

problems/0063.不同路径II.md

@@ -210,24 +210,54 @@ public:
 ```java
 class Solution {
     public int uniquePathsWithObstacles(int[][] obstacleGrid) {
-        int n = obstacleGrid.length, m = obstacleGrid[0].length;
+        int m = obstacleGrid.length;
-        int[][] dp = new int[n][m];
+        int n = obstacleGrid[0].length;
+        int[][] dp = new int[m][n];
 
-        for (int i = 0; i < m; i++) {
+        //如果在起点或终点出现了障碍，直接返回0
-	    if (obstacleGrid[0][i] == 1) break; //一旦遇到障碍，后续都到不了
+        if (obstacleGrid[m - 1][n - 1] == 1 || obstacleGrid[0][0] == 1) {
-	    dp[0][i] = 1;
+            return 0;
         }
-        for (int i = 0; i < n; i++) {
+
-	    if (obstacleGrid[i][0] == 1) break; ////一旦遇到障碍，后续都到不了
+        for (int i = 0; i < m && obstacleGrid[i][0] == 0; i++) {
             dp[i][0] = 1;
         }
-        for (int i = 1; i < n; i++) {
+        for (int j = 0; j < n && obstacleGrid[0][j] == 0; j++) {
-            for (int j = 1; j < m; j++) {
+            dp[0][j] = 1;
-                if (obstacleGrid[i][j] == 1) continue;
+        }
-                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
+
+        for (int i = 1; i < m; i++) {
+            for (int j = 1; j < n; j++) {
+                dp[i][j] = (obstacleGrid[i][j] == 0) ? dp[i - 1][j] + dp[i][j - 1] : 0;
             }
         }
-        return dp[n - 1][m - 1];
+        return dp[m - 1][n - 1];
+    }
+}
+```
+
+```java
+// 空间优化版本
+class Solution {
+    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
+        int m = obstacleGrid.length;
+        int n = obstacleGrid[0].length;
+        int[] dp = new int[n];
+        
+        for (int j = 0; j < n && obstacleGrid[0][j] == 0; j++) {
+            dp[j] = 1;
+        }
+        
+        for (int i = 1; i < m; i++) {
+            for (int j = 0; j < n; j++) {
+                if (obstacleGrid[i][j] == 1) {
+                    dp[j] = 0;
+                } else if (j != 0) {
+                    dp[j] += dp[j - 1];
+                }
+            }
+        }
+        return dp[n - 1];
     }
 }
 ```

problems/0070.爬楼梯.md

@@ -213,22 +213,6 @@ public:
 
 
 ### Java 
-```Java
-class Solution {
-    public int climbStairs(int n) {
-        // 跟斐波那契数列一样
-        if(n <= 2) return n;
-        int a = 1, b = 2, sum = 0;
-        
-        for(int i = 3; i <= n; i++){
-            sum = a + b;
-            a = b;
-            b = sum;
-        }
-        return b;
-    }
-}
-```
 
 ```java
 // 常规方式
@@ -241,15 +225,22 @@ public int climbStairs(int n) {
     }
     return dp[n];
 }
+```
+
+```Java
 // 用变量记录代替数组
-public int climbStairs(int n) {
+class Solution {
-    int a = 0, b = 1, c = 0; // 默认需要1次
+    public int climbStairs(int n) {
-    for (int i = 1; i <= n; i++) {
+        if(n <= 2) return n;
-        c = a + b;          // f(i - 1) + f(n - 2)
+        int a = 1, b = 2, sum = 0;
-        a = b;              // 记录上一轮的值
+        
-        b = c;              // 向后步进1个数
+        for(int i = 3; i <= n; i++){
+            sum = a + b;  // f(i - 1) + f(i - 2)
+            a = b;        // 记录f(i - 1)，即下一轮的f(i - 2)
+            b = sum;      // 记录f(i)，即下一轮的f(i - 1)
+        }
+        return b;
     }
-    return c;
 }
 ```
 

problems/0078.子集.md

@@ -180,10 +180,6 @@ class Solution {
     List<List<Integer>> result = new ArrayList<>();// 存放符合条件结果的集合
     LinkedList<Integer> path = new LinkedList<>();// 用来存放符合条件结果
     public List<List<Integer>> subsets(int[] nums) {
-        if (nums.length == 0){
-            result.add(new ArrayList<>());
-            return result;
-        }
         subsetsHelper(nums, 0);
         return result;
     }

problems/0112.路径总和.md

@@ -508,7 +508,7 @@ class solution:
 
             if cur_node.right: 
                 path.append(cur_node.right.val)
-                traversal(cur_node.right, remain-cur_node.left.val)
+                traversal(cur_node.right, remain-cur_node.right.val)
                 path.pop()
 
         result, path = [], []

problems/0135.分发糖果.md

@@ -135,25 +135,22 @@ class Solution {
          2、起点下标 ratings.length - 2 从右往左， 只要左边 比 右边 大，此时 左边的糖果应该 取本身的糖果数（符合比它左边大） 和 右边糖果数 + 1 二者的最大值，这样才符合 它比它左边的大，也比它右边大
     */
     public int candy(int[] ratings) {
-        int[] candyVec = new int[ratings.length];
+        int len = ratings.length;
+        int[] candyVec = new int[len];
         candyVec[0] = 1;
-        for (int i = 1; i < ratings.length; i++) {
+        for (int i = 1; i < len; i++) {
-            if (ratings[i] > ratings[i - 1]) {
+            candyVec[i] = (ratings[i] > ratings[i - 1]) ? candyVec[i - 1] + 1 : 1;
-                candyVec[i] = candyVec[i - 1] + 1;
-            } else {
-                candyVec[i] = 1;
-            }
         }
 
-        for (int i = ratings.length - 2; i >= 0; i--) {
+        for (int i = len - 2; i >= 0; i--) {
             if (ratings[i] > ratings[i + 1]) {
                 candyVec[i] = Math.max(candyVec[i], candyVec[i + 1] + 1);
             }
         }
 
         int ans = 0;
-        for (int s : candyVec) {
+        for (int num : candyVec) {
-            ans += s;
+            ans += num;
         }
         return ans;
     }

problems/0200.岛屿数量.广搜版.md

@@ -144,3 +144,50 @@ public:
 };
 
 ```
+
+## 其他语言版本
+
+### Java
+
+```java
+class Solution {
+
+    boolean[][] visited;
+    int[][] move = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
+
+    public int numIslands(char[][] grid) {
+        int res = 0;
+        visited = new boolean[grid.length][grid[0].length];
+        for(int i = 0; i < grid.length; i++) {
+            for(int j = 0; j < grid[0].length; j++) {
+                if(!visited[i][j] && grid[i][j] == '1') {
+                    bfs(grid, i, j);
+                    res++;
+                }
+            }
+        }
+        return res;
+    }
+
+    //将这片岛屿上的所有陆地都访问到
+    public void bfs(char[][] grid, int y, int x) {
+        Deque<int[]> queue = new ArrayDeque<>();
+        queue.offer(new int[]{y, x});
+        visited[y][x] = true;
+        while(!queue.isEmpty()) {
+            int[] cur = queue.poll();
+            int m = cur[0];
+            int n = cur[1];
+            for(int i = 0; i < 4; i++) {
+                int nexty = m + move[i][0];
+                int nextx = n + move[i][1];
+                if(nextx < 0 || nexty == grid.length || nexty < 0 || nextx == grid[0].length) continue;
+                if(!visited[nexty][nextx] && grid[nexty][nextx] == '1') {
+                    queue.offer(new int[]{nexty, nextx}); 
+                    visited[nexty][nextx] = true; //只要加入队列就标记为访问
+                }
+            }
+        }
+    }
+}
+```

problems/0216.组合总和III.md

@@ -260,6 +260,37 @@ class Solution {
 		}
 	}
 }
+
+// 上面剪枝 i <= 9 - (k - path.size()) + 1; 如果还是不清楚
+// 也可以改为 if (path.size() > k) return; 执行效率上是一样的
+class Solution {
+    LinkedList<Integer> path = new LinkedList<>();
+    List<List<Integer>> ans = new ArrayList<>();
+    public List<List<Integer>> combinationSum3(int k, int n) {
+        build(k, n, 1, 0);
+        return ans;
+    }
+
+    private void build(int k, int n, int startIndex, int sum) {
+        
+        if (sum > n) return;
+
+        if (path.size() > k) return;
+
+        if (sum == n && path.size() == k) {
+            ans.add(new ArrayList<>(path));
+            return;
+        }
+        
+        for(int i = startIndex; i <= 9; i++) {
+            path.add(i);
+            sum += i;
+            build(k, n, i + 1, sum);
+            sum -= i;
+            path.removeLast();
+        }
+    }
+}
 ```
 
 其他方法

problems/0337.打家劫舍III.md

@@ -191,7 +191,7 @@ public:
         vector<int> left = robTree(cur->left);
         vector<int> right = robTree(cur->right);
         // 偷cur，那么就不能偷左右节点。
-        int val1 = cur->val + left[1] + right[1];
+        int val1 = cur->val + left[0] + right[0];
         // 不偷cur，那么可以偷也可以不偷左右节点，则取较大的情况
         int val2 = max(left[0], left[1]) + max(right[0], right[1]);
         return {val2, val1};

problems/0376.摆动序列.md

@@ -269,18 +269,23 @@ class Solution:
 **贪心**
 ```golang
 func wiggleMaxLength(nums []int) int {
-	var count, preDiff, curDiff int //初始化默认为0
+	n := len(nums)
-	count = 1                       // 初始化为1，因为最小的序列是1个数
+	if n < 2 {
-	if len(nums) < 2 {
+		return n
-		return count
 	}
-	for i := 0; i < len(nums)-1; i++ {
+	ans := 1
-		curDiff = nums[i+1] - nums[i]
+	prevDiff := nums[1] - nums[0]
-		if (curDiff > 0 && preDiff <= 0) || (curDiff < 0 && preDiff >= 0) {
+	if prevDiff != 0 {
-			count++
+		ans = 2
+	}
+	for i := 2; i < n; i++ {
+		diff := nums[i] - nums[i-1]
+		if diff > 0 && prevDiff <= 0 || diff < 0 && prevDiff >= 0 {
+			ans++
+			prevDiff = diff
 		}
 	}
-	return count
+	return ans
 }
 ```
 

problems/0452.用最少数量的箭引爆气球.md

@@ -137,29 +137,21 @@ public:
 ### Java 
 ```java
 /**
-时间复杂度 : O(NlogN)  排序需要 O(NlogN) 的复杂度
+ * 时间复杂度 : O(NlogN)  排序需要 O(NlogN) 的复杂度
-
+ * 空间复杂度 : O(logN) java所使用的内置函数用的是快速排序需要 logN 的空间
-空间复杂度 : O(logN) java所使用的内置函数用的是快速排序需要 logN 的空间
+ */
-*/
 class Solution {
     public int findMinArrowShots(int[][] points) {
-        if (points.length == 0) return 0;
+        // 根据气球直径的开始坐标从小到大排序
-        //用x[0] - y[0] 会大于2147483647 造成整型溢出
+        // 使用Integer内置比较方法，不会溢出
-        Arrays.sort(points, (x, y) -> Integer.compare(x[0], y[0]));
+        Arrays.sort(points, (a, b) -> Integer.compare(a[0], b[0]));
-        //count = 1 因为最少需要一个箭来射击第一个气球
+
-        int count = 1;
+        int count = 1;  // points 不为空至少需要一支箭
-        //重叠气球的最小右边界
-        int leftmostRightBound  = points[0][1];
         for (int i = 1; i < points.length; i++) {
-        //如果下一个气球的左边界大于最小右边界
+            if (points[i][0] > points[i - 1][1]) {  // 气球i和气球i-1不挨着，注意这里不是>=
-        for(int i = 1; i < points.length; i++){
+                count++; // 需要一支箭
-            if (points[i][0] > leftmostRightBound ) {
+            } else {  // 气球i和气球i-1挨着
-                //增加一次射击
+                points[i][1] = Math.min(points[i][1], points[i - 1][1]); // 更新重叠气球最小右边界
-                count++;
-                leftmostRightBound  = points[i][1];
-                //不然就更新最小右边界
-            } else {
-                leftmostRightBound  = Math.min(leftmostRightBound , points[i][1]);
             }
         }
         return count;

problems/0463.岛屿的周长.md

@@ -179,6 +179,25 @@ class Solution:
 ```
 
 Go：
+```go
+func islandPerimeter(grid [][]int) int {
+    m, n := len(grid), len(grid[0])
+    res := 0
+    for i := 0; i < m; i++ {
+        for j := 0; j < n; j++ {
+            if grid[i][j] == 1 {
+                res += 4
+                // 上下左右四个方向
+                if i > 0 && grid[i-1][j] == 1 {res--} // 上边有岛屿
+                if i < m-1 && grid[i+1][j] == 1 {res--} // 下边有岛屿
+                if j > 0 && grid[i][j-1] == 1 {res--} // 左边有岛屿
+                if j < n-1 && grid[i][j+1] == 1 {res--} // 右边有岛屿
+            }
+        }
+    }
+    return res
+}
+```
 
 JavaScript：
 ```javascript

problems/0530.二叉搜索树的最小绝对差.md

@@ -233,7 +233,7 @@ class Solution:
             else: # 逐一处理节点
                 cur = stack.pop()
                 if pre: # 当前节点和前节点的值的差值
-                    result = min(result, cur.val - pre.val)
+                    result = min(result, abs(cur.val - pre.val))
                 pre = cur
                 cur = cur.right
         return result

problems/0695.岛屿的最大面积.md

@@ -167,3 +167,111 @@ public:
 };
 
 ```
+
+# 其它语言版本
+
+## Python
+### BFS
+```python
+class Solution:
+    def __init__(self):
+        self.count = 0
+
+    def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
+        # 与200.独立岛屿不同的是：此题grid列表内是int！！！
+
+        # BFS
+        if not grid: return 0
+
+        m, n = len(grid), len(grid[0])
+        visited = [[False for i in range(n)] for j in range(m)]
+
+        result = 0
+        for i in range(m):
+            for j in range(n):
+                if not visited[i][j] and grid[i][j] == 1:
+                    # 每一个新岛屿
+                    self.count = 0
+                    print(f'{self.count}')
+                    self.bfs(grid, visited, i, j)
+                    result = max(result, self.count)
+
+        return result
+
+    def bfs(self, grid, visited, i, j):
+        self.count += 1
+        visited[i][j] = True    
+
+        queue = collections.deque([(i, j)])
+        while queue:
+            x, y = queue.popleft()
+            for new_x, new_y in [(x + 1, y), (x - 1, y), (x, y - 1), (x, y + 1)]:
+                if 0 <= new_x < len(grid) and 0 <= new_y < len(grid[0]) and not visited[new_x][new_y] and grid[new_x][new_y] == 1:
+                    visited[new_x][new_y] = True
+                    self.count += 1
+                    queue.append((new_x, new_y))
+```
+### DFS
+```python
+class Solution:
+    def __init__(self):
+        self.count = 0
+
+    def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
+        # DFS
+        if not grid: return 0
+
+        m, n = len(grid), len(grid[0])
+        visited = [[False for _ in range(n)] for _ in range(m)]
+
+        result = 0
+        for i in range(m):
+            for j in range(n):
+                if not visited[i][j] and grid[i][j] == 1:
+                    self.count = 0
+                    self.dfs(grid, visited, i, j)
+                    result = max(result, self.count)
+        return result
+        
+    def dfs(self, grid, visited, x, y):
+        if visited[x][y] or grid[x][y] == 0:
+            return 
+        visited[x][y] = True
+        self.count += 1
+        for new_x, new_y in [(x + 1, y), (x - 1, y), (x, y + 1), (x, y - 1)]:
+            if 0 <= new_x < len(grid) and 0 <= new_y < len(grid[0]): 
+                self.dfs(grid, visited, new_x, new_y)
+```
+
+
+
+## Java
+
+这里使用深度优先搜索 DFS 来完成本道题目。我们使用 DFS 计算一个岛屿的面积，同时维护计算过的最大的岛屿面积。同时，为了避免对岛屿重复计算，我们在 DFS 的时候对岛屿进行 “淹没” 操作，即将岛屿所占的地方置为 0。
+
+```java
+public int maxAreaOfIsland(int[][] grid) {
+    int res = 0;
+    for(int i = 0;i < grid.length;i++){
+        for(int j = 0;j < grid[0].length;j++){
+        	//每遇到一个岛屿就计算这个岛屿的面积同时”淹没“这个岛屿
+            if(grid[i][j] == 1){
+            	//每次计算一个岛屿的面积都要与res比较，维护最大的岛屿面积作为最后的答案
+                res = Math.max(res,dfs(grid,i,j));
+            }
+        }
+    }
+    return res;
+}
+public int dfs(int[][] grid,int i,int j){
+	//搜索边界：i，j超过grid的范围或者当前元素为0，即当前所在的地方已经是海洋
+    if(i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == 0) return 0;
+    //淹没土地，防止后续被重复计算
+    grid[i][j] = 0;
+    //递归的思路：要求当前土地(i,j)所在的岛屿的面积，则等于1加上下左右相邻的土地的总面积
+    return 1 + dfs(grid,i - 1,j) +
+               dfs(grid,i + 1,j) +
+               dfs(grid,i,j + 1) +
+               dfs(grid,i,j - 1);
+}
+```

problems/0746.使用最小花费爬楼梯.md

@@ -66,7 +66,7 @@
 
 所以初始化代码为：
 
-```
+```CPP
 vector<int> dp(cost.size());
 dp[0] = cost[0];
 dp[1] = cost[1];
@@ -201,15 +201,32 @@ public:
 
 
 ### Java 
+
 ```Java
+// 方式一：第一步支付费用
 class Solution {
     public int minCostClimbingStairs(int[] cost) {
-        if (cost == null || cost.length == 0) {
+        int len = cost.length;
-            return 0;
+        int[] dp = new int[len + 1];
+
+        // 从下标为 0 或下标为 1 的台阶开始，因此支付费用为0
+        dp[0] = 0;
+        dp[1] = 0;
+
+        // 计算到达每一层台阶的最小费用
+        for (int i = 2; i <= len; i++) {
+            dp[i] = Math.min(dp[i - 1] + cost[i - 1], dp[i - 2] + cost[i - 2]);
         }
-        if (cost.length == 1) {
+
-            return cost[0];
+        return dp[len];
     }
+}
+```
+
+```Java
+// 方式二：第一步不支付费用
+class Solution {
+    public int minCostClimbingStairs(int[] cost) {
         int[] dp = new int[cost.length];
         dp[0] = cost[0];
         dp[1] = cost[1];

problems/0797.所有可能的路径.md

@@ -296,5 +296,29 @@ public:
 ### Java 
 
 ### Python
+```
+class Solution:
+    def __init__(self):
+        self.result = []
+        self.path = [0]
+
+    def allPathsSourceTarget(self, graph: List[List[int]]) -> List[List[int]]:
+        if not graph: return []
+
+        self.dfs(graph, 0)
+        return self.result
+    
+    def dfs(self, graph, root: int):
+        if root == len(graph) - 1:  # 成功找到一条路径时
+            # ***Python的list是mutable类型***
+            # ***回溯中必须使用Deep Copy***
+            self.result.append(self.path[:]) 
+            return
+        
+        for node in graph[root]:   # 遍历节点n的所有后序节点
+            self.path.append(node)
+            self.dfs(graph, node)
+            self.path.pop() # 回溯
+```
 
 ### Go 

problems/0968.监控二叉树.md

@@ -71,7 +71,7 @@
 
 后序遍历代码如下：
 
-```
+```CPP
 int traversal(TreeNode* cur) {
 
     // 空节点，该节点有覆盖
@@ -124,7 +124,7 @@ int traversal(TreeNode* cur) {
 
 代码如下：
 
-```
+```CPP
 // 空节点，该节点有覆盖
 if (cur == NULL) return 2;
 ```
@@ -143,7 +143,7 @@ if (cur == NULL) return 2;
 
 代码如下：
 
-```
+```CPP
 // 左右节点都有覆盖
 if (left == 2 && right == 2) return 0;
 ```
@@ -163,7 +163,7 @@ left == 2 && right == 0 左节点覆盖，右节点无覆盖
 此时摄像头的数量要加一，并且return 1，代表中间节点放摄像头。
 
 代码如下：
-```
+```CPP
 if (left == 0 || right == 0) {
     result++;
     return 1;
@@ -180,7 +180,7 @@ left == 1 && right == 1 左右节点都有摄像头
 
 代码如下：
 
-```
+```CPP
 if (left == 1 || right == 1) return 2;
 ```
 
@@ -198,7 +198,7 @@ if (left == 1 || right == 1) return 2;
 
 所以递归结束之后，还要判断根节点，如果没有覆盖，result++，代码如下：
 
-```
+```CPP
 int minCameraCover(TreeNode* root) {
     result = 0;
     if (traversal(root) == 0) { // root 无覆盖

problems/1356.根据数字二进制下1的数目排序.md

@@ -170,6 +170,39 @@ class Solution:
 ## Go
 
 ```go
+func sortByBits(arr []int) []int {
+    var tmp int
+    for i := 0; i < len(arr); i++ {
+        for j := i+1; j < len(arr); j++ {
+            // 冒泡排序的手法，但是排序的规则从比大小变成了比位运算1的个数
+            if isCmp(arr[i], arr[j]) {
+                tmp = arr[i]
+                arr[i] = arr[j]
+                arr[j] = tmp
+            }
+        } 
+    }
+    return arr
+}
+
+func isCmp(a, b int) bool {
+    bitA := bitCount(a)
+    bitB := bitCount(b)
+    if bitA == bitB {
+        return a > b
+    } else {
+        return bitA > bitB 
+    }
+}
+
+func bitCount(n int) int {
+    count := 0
+    for n != 0 {
+        n &= (n-1) // 清除最低位的1
+        count++
+    }
+    return count
+}
 ```
 
 ## JavaScript

problems/周总结/20201017二叉树周末总结.md

@@ -102,7 +102,7 @@
 
 现在已经讲过了几种二叉树了，二叉树，二叉平衡树，完全二叉树，二叉搜索树，后面还会有平衡二叉搜索树。 那么一些同学难免会有混乱了，我针对如下三个问题，帮大家在捋顺一遍：
 
-1. 平衡二叉搜索数是不是二叉搜索树和平衡二叉树的结合？
+1. 平衡二叉搜索树是不是二叉搜索树和平衡二叉树的结合？
 
 是的，是二叉搜索树和平衡二叉树的结合。