leetcode1319

class Solution:
    def bfs(self,connections,visited,l,neighbours):
        while len(l) > 0:
            temp = []
            while len(l) > 0:
                p = l.pop()
                if visited[p] == 0:
                    visited[p] = 1
                    if p in neighbours:
                        for q in neighbours[p]:
                            if visited[q] == 0:
                                temp.append(q)
            if len(temp) > 0:
                l = temp[:]


    def makeConnected(self, n: int, connections: 'List[List[int]]') -> int:
        m = len(connections)
        if n - m > 1:
            return -1#线不够用
        visited = [0] * n
        neighbours = {}
        for i in range(m):
            begin,end = connections[i][0],connections[i][1]
            if begin not in neighbours:
                neighbours[begin] = [end]
            else:
                neighbours[begin].append(end)
            if end not in neighbours:
                neighbours[end] = [begin]
            else:
                neighbours[end].append(begin)

        singleparts = 0
        for i in range(n):
            if visited[i] == 0:
                singleparts += 1
                self.bfs(connections,visited,[i],neighbours)

        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。