200. 岛屿数量（中）

• 题目
• 题解：DFS

题目

• 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

题解：DFS

• 遍历图中所有的位置，如果该位置为1，则向外扩展将与之相连的1全部变为0

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        m = len(grid)#行数
        n = len(grid[0])#列数
        res = 0
        def dfs(x, y):
            if grid[x][y] == '1':#本身改为0
                grid[x][y] = '0'
            else:
                return
            #相连的1都改为0
            if x > 0:#左边递归
                dfs(x - 1, y)
            if x < m - 1:#右边递归
                dfs(x + 1, y)
            if y > 0:#上边递归
                dfs(x, y - 1)
            if y < n - 1:#下边递归
                dfs(x, y + 1)
        for i in range(m):
            for j in range(n):
                if grid[i][j] == '1':
                    dfs(i, j)
                    res += 1#岛屿数加1
        return res