leetcode1267

Posted 2020-11-22 AsenYang

 1 class Solution:
 2     def dfs(self,grid,m,n,i,j):
 3         grid[i][j] = 0
 4         res = 1
 5         for x in range(m):
 6             if grid[x][j] == 1:
 7                 res += self.dfs(grid,m,n,x,j)
 8         for y in range(n):
 9             if grid[i][y] == 1:
10                 res += self.dfs(grid,m,n,i,y)
11         return res
12 
13     def countServers(self, grid: ‘List[List[int]]‘) -> int:
14         m = len(grid)#row
15         n = len(grid[0])#column
16         count = 0
17         direct = [[-1,0],[1,0],[0,-1],[0,1]]
18         for i in range(m):
19             for j in range(n):
20                 if grid[i][j] == 1:
21                     temp = self.dfs(grid,m,n,i,j)
22                     if temp > 1:
23                         count += temp
24         return count

这道题目使用DFS处理，但是dfs内部的逻辑不太好想，比赛的时候开始有思路，调试了几次都不对，脑子就变空白了，一直没有做出来。

参考：https://leetcode.com/problems/count-servers-that-communicate/discuss/436167/Simple-java-DFS-solution-similar-to-200.-number-of-islands