孤岛数 leetcode python

Question

问题如下：给定一个由“1”（陆地）和“0”（水）组成的二维网格图，计算岛屿的数量。岛屿四面环水，由相邻陆地水平或垂直连接而成。您可以假设网格的所有四个边缘都被水包围。

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        # use dfs / replace the island's elements with a sign "."
        a = len(grid)
        b = len(grid[0])
        num = 0


        for i in range(0 , a-1):
            for j in range(0 , b-1):
                if grid[i][j] == '1':
                    num += 1
                    self.dfs(grid , i , j)

        return num

    def dfs(self, grid , x , y):

        if x < 0 or y < 0 or x >= len(grid[0]) or y >= len(grid) or grid[x][y] == '0' or grid[x][y] == ".": # if out of boundary
            return

        grid[x][y] = "."        
        # if 1's has only one neighbor

        self.dfs(grid , x-1 , y)   # check 4 edges of water
        self.dfs(grid, x+1 , y)
        self.dfs(grid , x , y+1)
        self.dfs(grid , x , y- 1)

此代码仅适用于以下输出：[["1","1","1","1","0"],["1","1","0","1" ,"0"],["1","1","0","0","0"],["0","0","0","0","0"]]

但是，它不适用于此输出：[["1","1","0","0","0"],["1","1","0"," 0","0"],["0","0","1","0","0"],["0","0","0","1","1" ]]

我正在尝试使用 dfs 方法。因此，如果 x 在数组中，函数 dfs 将从 4 个边检查相邻元素是否为“1”，如果为“1”，则将其替换为“.”。否则，如果元素为“0”，则递归循环停止。一旦所有可能的组合的所有递归循环停止，计数就会增加 1。然后程序运行以查找其他“X”。

但是，该程序无法正常运行。因此，谁能帮我找出这段代码中的问题？

Answer 1

不幸的是，我找不到你的代码有什么问题，我自己做了一个，它有点不同但它有效，也许它可以帮助你找到问题

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        
        def is_land(grid, next_loc):
            
            (x, y) = next_loc
            
            grid[x][y] = '0'
            
            if x+1 < rows and grid[x+1][y] == '1':
                grid = is_land(grid, (x+1, y))
            
            if x - 1 >= 0 and grid[x-1][y] == '1':
                grid = is_land(grid, (x-1, y))
            
            if y+1 < cols and grid[x][y+1] == '1':
                grid = is_land(grid, (x, y+1))
            
            if y-1 >= 0 and grid[x][y-1] == '1':
                grid = is_land(grid, (x, y-1))
            
            return grid
        
        rows = len(grid)
        cols = len(grid[0])
        ans = 0
        for row in range(rows):
            for col in range(cols):
                if grid[row][col] == '1':
                    grid = is_land(grid, (row, col))
                    ans += 1
        return ans