力扣热题51

200. 岛屿数量 - 力扣（LeetCode）https://leetcode.cn/problems/number-of-islands/submissions/628256822/?envType=study-plan-v2&envId=top-100-liked

深度优先递归

双层循环进行遍历，对每个1进行计数作为岛屿，然后进入递归将这个1四周的1的四周都变为0即淹没，最后得到答案

#python
class Solution:
    def dfs(self,grid,a,b):
        grid[a][b]=0
        c,d=len(grid),len(grid[0])
        for x,y in [(a-1,b),(a+1,b),(a,b-1),(a,b+1)]:
            if 0<=x<c and 0<=y<d and grid[x][y]=="1":
                self.dfs(grid,x,y)
    def numIslands(self, grid: List[List[str]]) -> int:
        a=len(grid)
        if a==0:
            return 0
        b=len(grid[0])
        ans=0
        for i in range(a):
            for j in range(b):
                if grid[i][j]=="1":
                    ans=ans+1
                    self.dfs(grid,i,j)
        return ans