Update 0695.岛屿的最大面积.md

补充Python实现

Author: casnz1601

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);
 }
-```
+```