200. 岛屿数量

[题目链接](200. 岛屿数量 - 力扣（LeetCode）)

解题思路：用感染函数，遇到1，岛屿数目就加1，然后递归把上下左右变成2(以免下次遇到相同的岛屿)

代码

class Solution:
    
    # 感染函数，将[i, j]感染
    def process(self, grid, i, j):
        if i < 0 or i >= len(grid) or j < 0 or j >= len(grid[0]) or grid[i][j] != '1':
            return
        grid[i][j] = '2'
        self.process(grid, i - 1, j)
        self.process(grid, i + 1, j)
        self.process(grid, i, j - 1)
        self.process(grid, i, j + 1)
        
    
    def numIslands(self, grid: List[List[str]]) -> int:

        ans = 0
        for i in range(len(grid)):
            for j in range(len(grid[0])):
               if grid[i][j] == '1':
                   ans += 1
                   self.process(grid, i, j)
        return ans