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].

題目理解：

一共有N個人，相互討厭的人不能放在一組，問能否把所有人分成兩組

解題思路：

使用並查集，將相互討厭的人鏈接起來，如果出現長度為奇數的環，那麼就無法將這些人分成兩組

程式碼如下：

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);
    }
}