[pull] master from youngyangyang04:master

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/0695.岛屿的最大面积.md

@@ -170,6 +170,81 @@ 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。
@@ -199,4 +274,4 @@ public int dfs(int[][] grid,int i,int j){
                dfs(grid,i,j + 1) +
                dfs(grid,i,j - 1);
 }
-```
+```