[pull] master from youngyangyang04:master

problems/0039.组合总和.md

@@ -311,7 +311,7 @@ class Solution:
 
         for i in range(startIndex, len(candidates)):
             if total + candidates[i] > target:
-                break
+                continue
             total += candidates[i]
             path.append(candidates[i])
             self.backtracking(candidates, target, total, i, path, result)

problems/0200.岛屿数量.广搜版.md

@@ -197,6 +197,7 @@ class Solution {
 }
 ```
 
+
 ### Python
 BFS solution 
 ```python
@@ -236,6 +237,7 @@ class Solution:
                     continue
                 q.append((next_i, next_j))
                 visited[next_i][next_j] = True
+
 ```
 
 <p align="center">

problems/0200.岛屿数量.深搜版.md

@@ -218,6 +218,67 @@ class Solution {
     }
 }
 ```
+
+Python:
+
+```python
+# 版本一
+class Solution:
+    def numIslands(self, grid: List[List[str]]) -> int:
+        m, n = len(grid), len(grid[0])
+        visited = [[False] * n for _ in range(m)]
+        dirs = [(-1, 0), (0, 1), (1, 0), (0, -1)]  # 四个方向
+        result = 0
+
+        def dfs(x, y):
+            for d in dirs:
+                nextx = x + d[0]
+                nexty = y + d[1]
+                if nextx < 0 or nextx >= m or nexty < 0 or nexty >= n:  # 越界了，直接跳过
+                    continue
+                if not visited[nextx][nexty] and grid[nextx][nexty] == '1':  # 没有访问过的同时是陆地的
+                    visited[nextx][nexty] = True
+                    dfs(nextx, nexty)
+
+        for i in range(m):
+            for j in range(n):
+                if not visited[i][j] and grid[i][j] == '1':
+                    visited[i][j] = True
+                    result += 1  # 遇到没访问过的陆地，+1
+                    dfs(i, j)  # 将与其链接的陆地都标记上 true
+
+        return result
+```
+
+```python
+# 版本二
+class Solution:
+    def numIslands(self, grid: List[List[str]]) -> int:
+        m, n = len(grid), len(grid[0])
+        visited = [[False] * n for _ in range(m)]
+        dirs = [(-1, 0), (0, 1), (1, 0), (0, -1)]  # 四个方向
+        result = 0
+
+        def dfs(x, y):
+            if visited[x][y] or grid[x][y] == '0':
+                return  # 终止条件：访问过的节点 或者 遇到海水
+            visited[x][y] = True
+            for d in dirs:
+                nextx = x + d[0]
+                nexty = y + d[1]
+                if nextx < 0 or nextx >= m or nexty < 0 or nexty >= n:  # 越界了，直接跳过
+                    continue
+                dfs(nextx, nexty)
+
+        for i in range(m):
+            for j in range(n):
+                if not visited[i][j] and grid[i][j] == '1':
+                    result += 1  # 遇到没访问过的陆地，+1
+                    dfs(i, j)  # 将与其链接的陆地都标记上 true
+
+        return result
+```
+
 <p align="center">
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">
   <img src="../pics/网站星球宣传海报.jpg" width="1000"/>

problems/0674.最长连续递增序列.md

@@ -302,8 +302,9 @@ func findLengthOfLCIS(nums []int) int {
 }
 ```
 
-### Rust:
 
+### Rust:
+>动态规划
 ```rust
 pub fn find_length_of_lcis(nums: Vec<i32>) -> i32 {
     if nums.is_empty() {
@@ -321,6 +322,27 @@ pub fn find_length_of_lcis(nums: Vec<i32>) -> i32 {
 }
 ```
 
+
+> 贪心
+
+```rust
+impl Solution {
+    pub fn find_length_of_lcis(nums: Vec<i32>) -> i32 {
+        let (mut res, mut count) = (1, 1);
+        for i in 1..nums.len() {
+            if nums[i] > nums[i - 1] {
+                count += 1;
+                res = res.max(count);
+                continue;
+            }
+            count = 1;
+        }
+        res
+    }
+}
+```
+
+
 ### Javascript：
 
 > 动态规划：