遍历二维数组,找到为 1 的之后,使用深度优先搜索将与其相邻的陆地全部淹没(也就是1 改为 0)

#
# 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
#
# 判断岛屿数量
# @param grid char字符型二维数组 
# @return int整型
#
class Solution:

    def solve(self , grid: List[List[str]]) -> int:
        # write code here
        m=len(grid)
        n=len(grid[0])
        num=0
        def dfs(grid, x, y):
            # 先“淹掉”自己
            grid[x][y] = '0'
            if x + 1 < m and grid[x+1][y] == '1': dfs(grid,x+1, y)
            if x - 1 >= 0 and grid[x-1][y] == '1': dfs(grid,x-1, y)
            if y + 1 < n and grid[x][y+1] == '1': dfs(grid,x, y+1)
            if y - 1 >= 0 and grid[x][y-1] == '1': dfs(grid,x, y-1)
        for i in range(m):
            for j in range(n):
                if grid[i][j]=='1':
                    dfs(grid,i,j)
                    num+=1
        return num