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

Author: youngyangyang04

problems/0024.两两交换链表中的节点.md

@@ -409,7 +409,7 @@ impl Solution {
 // 递归
 impl Solution {
     pub fn swap_pairs(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
-        if head == None || head.as_ref().unwrap().next == None {
+        if head.is_none() || head.as_ref().unwrap().next.is_none() {
             return head;
         }
 

problems/0034.在排序数组中查找元素的第一个和最后一个位置.md

@@ -240,7 +240,7 @@ class Solution {
 		while (left - 1 >= 0 && nums[left - 1] == nums[index]) { // 防止数组越界。逻辑短路，两个条件顺序不能换
 			left--;
 		}
-        // 向左滑动，找右边界
+        // 向右滑动，找右边界
 		while (right + 1 < nums.length && nums[right + 1] == nums[index]) { // 防止数组越界。
 			right++;
 		}

problems/0047.全排列II.md

@@ -98,8 +98,10 @@ public:
     }
 };
 
+// 时间复杂度: 最差情况所有元素都是唯一的。复杂度和全排列1都是 O(n! * n) 对于 n 个元素一共有 n! 中排列方案。而对于每一个答案，我们需要 O(n) 去复制最终放到 result 数组
+// 空间复杂度: O(n) 回溯树的深度取决于我们有多少个元素
 ```
-* 时间复杂度: O(n)
+* 时间复杂度: O(n! * n)
 * 空间复杂度: O(n)
 
 ## 拓展

problems/0063.不同路径II.md

@@ -397,7 +397,39 @@ class Solution:
 
 ```
 
+动态规划（版本五）
 
+```python
+class Solution:
+    def uniquePathsWithObstacles(self, obstacleGrid):
+        if obstacleGrid[0][0] == 1:
+            return 0
+        
+        m, n = len(obstacleGrid), len(obstacleGrid[0])
+        
+        dp = [0] * n  # 创建一个一维列表用于存储路径数
+        
+        # 初始化第一行的路径数
+        for j in range(n):
+            if obstacleGrid[0][j] == 1:
+                break
+            dp[j] = 1
+
+        # 计算其他行的路径数
+        for i in range(1, m):
+            if obstacleGrid[i][0] == 1:
+                dp[0] = 0
+            for j in range(1, n):
+                if obstacleGrid[i][j] == 1:
+                    dp[j] = 0
+                    continue
+                
+                dp[j] += dp[j - 1]
+        
+        return dp[-1]  # 返回最后一个元素，即终点的路径数
+
+
+```
 ### Go
 
 ```go

problems/0070.爬楼梯完全背包版本.md

@@ -133,12 +133,13 @@ Java：
 class Solution {
     public int climbStairs(int n) {
         int[] dp = new int[n + 1];
-        int m = 2;
+        int m = 2; //有兩個物品：itme1重量爲一，item2重量爲二
         dp[0] = 1;
 
         for (int i = 1; i <= n; i++) { // 遍历背包
             for (int j = 1; j <= m; j++) { //遍历物品
-                if (i >= j) dp[i] += dp[i - j];
+                if (i >= j)  //當前的背包容量 大於 物品重量的時候，我們才需要記錄當前的這個裝得方法（方法數+）
+			dp[i] += dp[i - j];
             }
         }
 

problems/0072.编辑距离.md

@@ -40,6 +40,8 @@ exection -> execution (插入 'u')
 * 0 <= word1.length, word2.length <= 500
 * word1 和 word2 由小写英文字母组成
 
+# 算法公开课
+**《代码随想录》算法视频公开课：[动态规划终极绝杀！ LeetCode：72.编辑距离](https://www.bilibili.com/video/BV1we4y157wB/)，相信结合视频再看本篇题解，更有助于大家对本题的理解**。
 
 ## 思路
 

problems/0121.买卖股票的最佳时机.md

@@ -243,8 +243,27 @@ class Solution {
     }
 }
 ```
+> 动态规划：版本二(使用二維數組（和卡哥思路一致），下面還有使用一維滾動數組的更優化版本)
 
-> 动态规划：版本二
+```Java
+class Solution {
+    public int maxProfit(int[] prices) {
+        int len = prices.length;
+        int dp[][] = new int[2][2];
+        
+        dp[0][0] = - prices[0];
+        dp[0][1] = 0;
+
+        for (int i = 1; i < len; i++){
+            dp[i % 2][0] = Math.max(dp[(i - 1) % 2][0], - prices[i]);
+            dp[i % 2][1] = Math.max(dp[(i - 1) % 2][1], prices[i] + dp[(i - 1) % 2][0]);
+        }
+        return dp[(len - 1) % 2][1];
+    }
+}
+```
+
+> 动态规划：版本二(使用一維數組)
 
 ``` java
 class Solution {
@@ -271,6 +290,10 @@ class Solution {
   }
 }
 ```
+```Java
+
+```
+
 
 Python:
 
@@ -510,7 +533,37 @@ public class Solution
 }
 ```
 
+Rust:
+
+> 贪心
+
+```rust
+impl Solution {
+    pub fn max_profit(prices: Vec<i32>) -> i32 {
+        let (mut low, mut res) = (i32::MAX, 0);
+        for p in prices {
+            low = p.min(low);
+            res = res.max(p - low);
+        }
+        res
+    }
+}
+```
 
+> 动态规划
+
+```rust
+impl Solution {
+    pub fn max_profit(prices: Vec<i32>) -> i32 {
+        let mut dp = vec![-prices[0], 0];
+        for p in prices {
+            dp[0] = dp[0].max(-p);
+            dp[1] = dp[1].max(dp[0] + p);
+        }
+        dp[1]
+    }
+}
+```
 
 
 <p align="center">

problems/0122.买卖股票的最佳时机II.md

@@ -322,13 +322,10 @@ function maxProfit(prices: number[]): number {
 
 ```Rust
 impl Solution {
-    fn max(a: i32, b: i32) -> i32 {
-        if a > b { a } else { b }
-    }
     pub fn max_profit(prices: Vec<i32>) -> i32 {
         let mut result = 0;
         for i in 1..prices.len() {
-            result += Self::max(prices[i] - prices[i - 1], 0);
+            result += (prices[i] - prices[i - 1]).max(0);
         }
         result
     }
@@ -339,18 +336,14 @@ impl Solution {
 
 ```Rust
 impl Solution {
-    fn max(a: i32, b: i32) -> i32 {
-        if a > b { a } else { b }
-    }
     pub fn max_profit(prices: Vec<i32>) -> i32 {
-        let n = prices.len();
-        let mut dp = vec![vec![0; 2]; n];
-        dp[0][0] -= prices[0];
-        for i in 1..n {
-            dp[i][0] = Self::max(dp[i - 1][0], dp[i - 1][1] - prices[i]);
-            dp[i][1] = Self::max(dp[i - 1][1], dp[i - 1][0] + prices[i]);
+        let mut dp = vec![vec![0; 2]; prices.len()];
+        dp[0][0] = -prices[0];
+        for i in 1..prices.len() {
+            dp[i][0] = dp[i - 1][0].max(dp[i - 1][1] - prices[i]);
+            dp[i][1] = dp[i - 1][1].max(dp[i - 1][0] + prices[i]);
         }
-        Self::max(dp[n - 1][0], dp[n - 1][1])
+        dp[prices.len() - 1][1]
     }
 }
 ```

problems/0122.买卖股票的最佳时机II（动态规划）.md

@@ -154,7 +154,24 @@ class Solution
     }
 }
 ```
+```java
+//DP using 2*2 Array (下方還有使用一維滾動數組的更優化版本)
+class Solution {
+    public int maxProfit(int[] prices) {
+        int dp[][] = new int [2][2];
+        //dp[i][0]: holding the stock
+        //dp[i][1]: not holding the stock
+        dp[0][0] = - prices[0];
+        dp[0][1] = 0;
 
+        for(int i = 1; i < prices.length; i++){
+            dp[i % 2][0] = Math.max(dp[(i - 1) % 2][0], dp[(i - 1) % 2][1] - prices[i]);
+            dp[i % 2][1] = Math.max(dp[(i - 1) % 2][1], dp[(i - 1) % 2][0] + prices[i]);
+        }
+        return dp[(prices.length - 1) % 2][1];
+    }
+}
+```
 ```java
 // 优化空间
 class Solution {
@@ -346,7 +363,52 @@ public class Solution
 }
 ```
 
+Rust:
+
+> 贪心
 
+```rust
+impl Solution {
+    pub fn max_profit(prices: Vec<i32>) -> i32 {
+        let mut result = 0;
+        for i in 1..prices.len() {
+            result += (prices[i] - prices[i - 1]).max(0);
+        }
+        result
+    }
+}
+```
+
+>动态规划
+
+```rust
+impl Solution {
+    pub fn max_profit(prices: Vec<i32>) -> i32 {
+        let mut dp = vec![vec![0; 2]; prices.len()];
+        dp[0][0] = -prices[0];
+        for i in 1..prices.len() {
+            dp[i][0] = dp[i - 1][0].max(dp[i - 1][1] - prices[i]);
+            dp[i][1] = dp[i - 1][1].max(dp[i - 1][0] + prices[i]);
+        }
+        dp[prices.len() - 1][1]
+    }
+}
+```
+
+> 优化
+
+```rust
+impl Solution {
+    pub fn max_profit(prices: Vec<i32>) -> i32 {
+        let mut dp = vec![-prices[0], 0];
+        for p in prices {
+            dp[0] = dp[0].max(dp[1] - p);
+            dp[1] = dp[1].max(dp[0] + p);
+        }
+        dp[1]
+    }
+}
+```
 
 <p align="center">
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">

problems/0130.被围绕的区域.md

@@ -188,6 +188,54 @@ class Solution {
     }
 }
 ```
+```Java
+//BFS（使用helper function）
+class Solution {
+    int[][] dir ={{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
+    public void solve(char[][] board) {
+        for(int i = 0; i < board.length; i++){
+            if(board[i][0] == 'O') bfs(board, i, 0);
+            if(board[i][board[0].length - 1] == 'O') bfs(board, i, board[0].length - 1);
+        }
+
+        for(int j = 1 ; j < board[0].length - 1; j++){
+            if(board[0][j] == 'O') bfs(board, 0, j);
+            if(board[board.length - 1][j] == 'O') bfs(board, board.length - 1, j);
+        }
+
+        for(int i = 0; i < board.length; i++){
+            for(int j = 0; j < board[0].length; j++){
+                if(board[i][j] == 'O') board[i][j] = 'X';
+                if(board[i][j] == 'A') board[i][j] = 'O';
+            }
+        }
+    }
+    private void bfs(char[][] board, int x, int y){
+        Queue<Integer> que = new LinkedList<>();
+        board[x][y] = 'A';
+        que.offer(x);
+        que.offer(y);
+        
+        while(!que.isEmpty()){
+            int currX = que.poll();
+            int currY = que.poll();
+
+            for(int i = 0; i < 4; i++){
+                int nextX = currX + dir[i][0];
+                int nextY = currY + dir[i][1];
+
+            if(nextX < 0 || nextY < 0 || nextX >= board.length || nextY >= board[0].length)
+                continue;
+            if(board[nextX][nextY] == 'X'|| board[nextX][nextY] == 'A')
+                continue;
+            bfs(board, nextX, nextY);
+            }
+        }
+    }
+}
+
+```
+
 ```Java
 // 深度优先遍历
 // 使用 visited 数组进行标记
@@ -296,6 +344,47 @@ class Solution {
     }
 }
 ```
+```java
+//DFS(有終止條件)
+class Solution {
+    int[][] dir ={{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
+    public void solve(char[][] board) {
+
+        for(int i = 0; i < board.length; i++){
+            if(board[i][0] == 'O') dfs(board, i, 0);
+            if(board[i][board[0].length - 1] == 'O') dfs(board, i, board[0].length - 1);
+        }
+
+        for(int j = 1 ; j < board[0].length - 1; j++){
+            if(board[0][j] == 'O') dfs(board, 0, j);
+            if(board[board.length - 1][j] == 'O') dfs(board, board.length - 1, j);
+        }
+
+        for(int i = 0; i < board.length; i++){
+            for(int j = 0; j < board[0].length; j++){
+                if(board[i][j] == 'O') board[i][j] = 'X';
+                if(board[i][j] == 'A') board[i][j] = 'O';
+            }
+        }        
+    }
+
+    private void dfs(char[][] board, int x, int y){
+        if(board[x][y] == 'X'|| board[x][y] == 'A')
+            return;
+        board[x][y] = 'A';
+        for(int i = 0; i < 4; i++){
+            int nextX = x + dir[i][0];
+            int nextY = y + dir[i][1];
+            
+            if(nextX < 0 || nextY < 0 || nextX >= board.length || nextY >= board[0].length)
+                continue;
+            // if(board[nextX][nextY] == 'X'|| board[nextX][nextY] == 'A')
+            //     continue;
+            dfs(board, nextX, nextY);
+        }
+    }
+}
+```
 
 <p align="center">
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">