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 <= N <= 2000
0 <= dislikes.length <= 10000
1 <= dislikes[i][j] <= N
dislikes[i][0] < dislikes[i][1]
- There does not exist
i != j
for whichdislikes[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);
}
}