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

 

Constraints:

• 1 <= n <= 2000
• 0 <= dislikes.length <= 10000
• dislikes[i].length == 2
• 1 <= dislikes[i][j] <= n
• dislikes[i][0] < dislikes[i][1]
• There does not exist i != j for which dislikes[i] == dislikes[j].

class Solution {
    public boolean possibleBipartition(int n, int[][] dis) {
        int[] visited = new int[n + 1];
        List<Integer>[] graph = new ArrayList[n + 1];
        for(int i = 0; i <= n; i++) graph[i] = new ArrayList();
        for(int[] dislike: dis) {
            int fr = dislike[0], to = dislike[1];
            graph[fr].add(to);
            graph[to].add(fr);
        }
        for(int i = 1; i <= n; i++) {
            if(visited[i] == 0 && graph[i].size() > 0) {
                visited[i] = 1;
                Queue<Integer> q = new LinkedList();
                q.offer(i);
                while(!q.isEmpty()) {
                    int cur = q.poll();
                    for(int j : graph[cur]) {
                        if(visited[j] == 0) {
                            visited[j] = (visited[cur] == 1 ? 2 : 1);
                            q.offer(j);
                        }
                        else {
                            if(visited[j] == visited[cur]) return false;
                        }
                    }
                }
            }
        }
        return true;
    }
}

和785一样，但是得先建立graph，完了之后bfs涂色

class Solution {
    public boolean possibleBipartition(int N, int[][] dislikes) {
        Map<Integer, List<Integer>> map = new HashMap();
        for(int[] dis : dislikes) {
            map.computeIfAbsent(dis[0], a -> new ArrayList()).add(dis[1]);
            map.computeIfAbsent(dis[1], a -> new ArrayList()).add(dis[0]);
        }
        int[] color = new int[N + 1];
        
        for(int i = 1; i <= N; i++) {
            if(color[i] == 0) {
                color[i] = 1;
                Queue<Integer> q = new LinkedList();
                q.offer(i);
                while(!q.isEmpty()) {
                    int cur = q.poll();
                    if(map.containsKey(cur)) {
                        for(int nei : map.get(cur)) {
                            if(color[nei] == 0){
                                color[nei] = color[cur] == 1 ? 2 : 1;
                                q.offer(nei);
                            } 
                            else {
                                if(color[nei] == color[cur]) return false;
                            }
                            
                        }
                    }
                    
                }
            }
        }
        return true;
    }
}

 

标签：cur,int,dislikes,Possible,Bipartition,visited,new,dis,886
来源： https://www.cnblogs.com/wentiliangkaihua/p/14957132.html