leetcode 200. Number of Islands-优快云博客

博客围绕计算二维网格中岛屿数量的问题展开，介绍用Python重新实现BFS算法来解决该问题。指出在写BFS时可新建列表，用元组表示位置下标添加到列表。还提醒不能用grid[r, c] == '0'这种适用于NumPy数组的写法，具体思路可查看博客园链接。

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Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

Input:
11110
11010
11000
00000

Output: 1

Example 2:

Input:
11000
11000
00100
00011

Output: 3

具体的思路可以见我的博客园的博客：https://www.cnblogs.com/Weixu-Liu/p/10712642.html

在这个题中，我用python重新写BFS，在写BFS中，可以新建立一个列表，把位置下标用元组表示，加到列表中就行。

python代码：

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        sum = 0
        q = []
        m = len(grid)
        if m == 0:
            return 0
        n = len(grid[0])
        dx = [0,0,1,-1]
        dy = [1,-1,0,0]
        for r in range(m):
            for c in range(n):
                if grid[r][c] == '1':
                    grid[r][c] == '0'
                    q.append((r,c))
                    while len(q):
                        Q = q.pop(0)
                        index_r = Q[0]
                        index_c = Q[1]
                        for i in range(4):
                            x = index_r + dx[i]
                            y = index_c + dy[i]
                            if x >= 0 and x < m and y >= 0 and y < n and grid[x][y] == '1':
                                grid[x][y] = '0'
                                q.append((x,y))
                    sum += 1
        return sum