leetcode_200. 岛屿数量

给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。

岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。

此外，你可以假设该网格的四条边均被水包围。

 

示例 1：

输入：grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
输出：1
示例 2：

输入：grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
输出：3
 

提示：

m == grid.length
n == grid[i].length
1 <= m, n <= 300
grid[i][j] 的值为 '0' 或 '1'

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/number-of-islands
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        def dfs(grid,r,c):
            grid[r][c]=0
            nr=len(grid)
            nc=len(grid[0])
            for x, y in [(r-1,c),(r+1,c),(r,c-1),(r,c+1)]:
                if 0<=x<nr and 0<=y<nc and grid[x][y]=='1':
                    dfs(grid,x,y)
        
        nr=len(grid)
        if nr==0:return 0
        nc=len(grid[0])
        res=0
        for i in range(nr):
            for j in range(nc):
                if grid[i][j]=='1':
                    res+=1
                    dfs(grid,i,j)
        
        return res