200. 岛屿数量

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

这道题蛮重要的，频率蛮高，做个标记吧，dfs就行，遍历过的话就将这个设置为0

class Solution:

    def numIslands(self, grid: List[List[str]]) -> int:

        clen=len(grid)

        llen=len(grid[0])

        def dfs(grid,i,j):

            if not 0<=i<clen or not 0<=j<llen or grid[i][j]=='0':

                return

            grid[i][j]='0'

            dfs(grid,i+1,j)

            dfs(grid,i-1,j)

            dfs(grid,i,j+1)

            dfs(grid,i,j-1)

        res=0

        for i in range(clen):

            for j in range(llen):

                if grid[i][j]=='1':

                    dfs(grid,i,j)

                    res+=1

        return res

后面进行了一个剪枝，就是先判断能不能去，不为1的就不去。

优化后大概是优化了1/3这样，还不错的