Possible Bipartition

Given a set ofN people (numbered1, 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, ifdislikes[i] = [a, b], it means it is not allowed to put the people numbered aand binto the same group.

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

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 which dislikes[i] == dislikes[j].

Example

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

相比之前拿道题，多了一个建图的预处理

Code

class Solution {
    public boolean possibleBipartition(int N, int[][] dislikes) {
        List<List<Integer>> graph = new ArrayList<>();
        for (int i = 0; i <= N; i++) {
            graph.add(new ArrayList<>());
        }
        for (int[] elem : dislikes) {
            graph.get(elem[0]).add(elem[1]);
            graph.get(elem[1]).add(elem[0]);
        }

        int[] color = new int[N + 1];
        Queue<Integer> q = new LinkedList<>();
        //Divide the node in two sets such that no two nodes in same set are linked
        //[[1,2,3], [0,2], [0,1,3], [0,2]]
        for (int i = 1; i <= N; i++) {
            if (color[i] == 0) {
                q.offer(i);
                color[i] = 1;
                while (!q.isEmpty()) {
                    int node = q.poll();
                    for (int child : graph.get(node)) {
                        if (color[child] == 0) {
                            if (color[node] == 2) {
                                color[child] = 1;
                            } else {
                                color[child] = 2;
                            }
                            q.offer(child);
                        } else if (color[child] == color[node]) {
                            return false;
                        }                  
                    }
                }
            }
        }
        return true;
    }
}