leetcode算法题基础（二十五）深度优先（三）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:
        m,n=len(grid), len(grid[0])
        visited=[[0 for i in range(n)] for j in range(m)]
        res=0
        for i in range(m):
            for j in range(n):
                if visited[i][j]==0 and grid[i][j]=='1':
                    res+=1
                    self.dfs(m, n, i, j, grid, visited)
        return res

    def dfs(self, m, n, i, j, grid, visited):
        tmp=[[i,j]]
        visited[i][j]=1
        while tmp:
            i,j=tmp.pop()
            if i+1<m and visited[i+1][j]==0 and grid[i+1][j]=='1':
                visited[i+1][j]=1
                tmp.append([i+1, j])
            if j+1<n and visited[i][j+1]==0 and grid[i][j+1]=='1':
                visited[i][j+1]=1
                tmp.append([i, j+1])
            if i-1>=0 and visited[i-1][j]==0 and grid[i-1][j]=='1':
                visited[i-1][j]=1
                tmp.append([i-1, j])
            if j-1>=0 and visited[i][j-1]==0 and grid[i][j-1]=='1':
                visited[i][j-1]=1
                tmp.append([i, j-1])