200. 岛屿数量

1. [题目链接](200. 岛屿数量 - 力扣（LeetCode）)

2. 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)

3. 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