【LeetCode】886. Possible Bipartition 解题报告（Python）

【LeetCode】886. Possible Bipartition 解题报告（Python）

作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

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题目地址：https://leetcode.com/problems/possible-bipartition/description/

题目描述：

Given a set of N people (numbered 1, 2, ..., N), we would like to split everyone into two groups of any size.

Each person may dislike some other people, and they should not go into the same group.

Formally, if dislikes[i] = [a, b], it means it is not allowed to put the people numbered a and b into the same group.

Return true if and only if it is possible to split everyone into two groups in this way.

Example 1:

Input: N = 4, dislikes = [[1,2],[1,3],[2,4]]
Output: true
Explanation: group1 [1,4], group2 [2,3]

Example 2:

Input: N = 3, dislikes = [[1,2],[1,3],[2,3]]
Output: false

Example 3:

Input: N = 5, dislikes = [[1,2],[2,3],[3,4],[4,5],[1,5]]
Output: false

Note:

1. 1 <= N <= 2000
2. 0 <= dislikes.length <= 10000
3. 1 <= dislikes[i][j] <= N
4. dislikes[i][0] < dislikesi
5. There does not exist i != j for which dislikes[i] == dislikes[j].

题目大意

一群人中有些人不喜欢对方因此不能放到同一个组里，问所有的人能否划分成两个组。

解题方法

这个题还是要抽象出来，抽象出一个二分图的模型。即不喜欢对方的两个人属于二分图中不同的部分。所以，这个题和785. Is Graph Bipartite?一模一样的。

同样使用dfs去做，需要把每个节点都当做起始节点去染色，这样判断是否有冲突。染色的方式是0-未染色，1-染了红色，-1代表染了蓝色。

时间复杂度是O(V + E)，空间复杂度是O(V + E).

代码如下：

class Solution(object):
    def possibleBipartition(self, N, dislikes):
        """
        :type N: int
        :type dislikes: List[List[int]]
        :rtype: bool
        """
        graph = collections.defaultdict(list)
        for dislike in dislikes:
            graph[dislike[0] - 1].append(dislike[1] - 1)
            graph[dislike[1] - 1].append(dislike[0] - 1)
        color = [0] * N
        for i in range(N):
            if color[i] != 0: continue
            bfs = collections.deque()
            bfs.append(i)
            color[i] = 1
            while bfs:
                cur = bfs.popleft()
                for e in graph[cur]:
                    if color[e] != 0:
                        if color[cur] == color[e]:
                            return False
                    else:
                        color[e] = -color[cur]
                        bfs.append(e)
        return True

参考资料：

https://www.youtube.com/watch?v=VlZiMD7Iby4

日期

2018 年 9 月 24 日 —— 祝大家中秋节快乐