886. 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].

同 785. Is Graph Bipartite?　　https://www.cnblogs.com/fatttcat/p/10064006.html

Graph Coloring + DFS

先根据dislikes中的信息建graph，再dfs判断color

time: O(V+E), space: O(V+E)

class Solution {
    int[] colors;
    public boolean possibleBipartition(int N, int[][] dislikes) {
        List<Integer>[] g = new ArrayList[N];
        for(int i = 0; i < N; i++)
            g[i] = new ArrayList<>(N);
        
        for(int[] d : dislikes) {
            int a = d[0] - 1;
            int b = d[1] - 1;
            g[a].add(b);
            g[b].add(a);
        }
        
        colors = new int[N];
        for(int i = 0; i < N; i++) {
            if(colors[i] == 0 && !checkColor(g, i, 1))
                return false;
        }
        return true;
    }
    
    private boolean checkColor(List<Integer>[] graph, int node, int color) {
        colors[node] = color;
        for(int i = 0; i < graph[node].size(); i++) {
            int dis = graph[node].get(i);
            if(colors[dis] == colors[node])
                return false;
            if(colors[dis] == 0 && !checkColor(graph, dis, -color))
                return false;
        }
        return true;
    }
}