leetcode1319

 1 class Solution:
 2     def bfs(self,connections,visited,l,neighbours):
 3         while len(l) > 0:
 4             temp = []
 5             while len(l) > 0:
 6                 p = l.pop()
 7                 if visited[p] == 0:
 8                     visited[p] = 1
 9                     if p in neighbours:
10                         for q in neighbours[p]:
11                             if visited[q] == 0:
12                                 temp.append(q)
13             if len(temp) > 0:
14                 l = temp[:]
15 
16 
17     def makeConnected(self, n: int, connections: 'List[List[int]]') -> int:
18         m = len(connections)
19         if n - m > 1:
20             return -1#线不够用
21         visited = [0] * n
22         neighbours = {}
23         for i in range(m):
24             begin,end = connections[i][0],connections[i][1]
25             if begin not in neighbours:
26                 neighbours[begin] = [end]
27             else:
28                 neighbours[begin].append(end)
29             if end not in neighbours:
30                 neighbours[end] = [begin]
31             else:
32                 neighbours[end].append(begin)
33 
34         singleparts = 0
35         for i in range(n):
36             if visited[i] == 0:
37                 singleparts += 1
38                 self.bfs(connections,visited,[i],neighbours)
39 
40         return singleparts - 1

算法思想：BFS。

如果总的线条数量少于n-1，那么不可能将所有的节点连接在一个图中返回-1。

使用BFS遍历所有节点，计算不相连的子图个数，结果等于不相连的子图个数-1。

posted on 2020-01-13 14:22 Sempron2800+ 阅读(199) 评论(0) 编辑收藏举报