力扣：200. 岛屿数量（Python3）

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。

来源：力扣（LeetCode）
链接：力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台

示例：

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

解法：

遍历每个点，如果为1，计数+1并把周围相邻的1标记为2。

代码：

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        count = 0

        def infect(r, c):
            if 0 <= r < len(grid) and 0 <= c < len(grid[0]) and grid[r][c] == '1':
                grid[r][c] = '2'
                infect(r + 1, c)
                infect(r - 1, c)
                infect(r, c + 1)
                infect(r, c - 1)

        for idr, r in enumerate(grid):
            for idc, c in enumerate(r):
                if c == '1':
                    infect(idr, idc)
                    count += 1
        return count