Possible Bipartition

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] < dislikes[i][1]
5. There does not exist i != j for which dislikes[i] == dislikes[j].

题目理解：

一共有N个人，相互讨厌的人不能放在一组，问能否把所有人分成两组

解题思路：

使用并查集，将相互讨厌的人链接起来，如果出现长度为奇数的环，那么就无法将这些人分成两组

代码如下：

class Solution {
    public boolean possibleBipartition(int N, int[][] dislikes) {
    	int[] head = new int[N + 1];
    	for(int i = 0; i < N + 1; i++) {
    		head[i] = i;
    	}
    	for(int[] it : dislikes) {
    		int[] head_a = find(it[0], head, 0);
    		int[] head_b = find(it[1], head, 0);
    		if(head_a[0] == head_b[0] && (head_a[1] + head_b[1]) % 2 == 0)
    			return false;
    		head[it[1]] = it[0];
    	}
    	return true;
    }
    
    public int[] find(int cur, int[] head, int ct) {
    	if(cur == head[cur])
    		return new int[] {cur, ct};
    	return find(head[cur], head, ct + 1);
    }
}