力扣labuladong——一刷day79

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前言

给你一幅「图」，请你用两种颜色将图中的所有顶点着色，且使得任意一条边的两个端点的颜色都不相同，你能做到吗？ 这就是图的「双色问题」，其实这个问题就等同于二分图的判定问题，如果你能够成功地将图染色，那么这幅图就是一幅二分图，反之则不是：

一、力扣785. 判断二分图

class Solution {
    boolean ok = true;
    boolean[] visited;
    boolean[] color;
    public boolean isBipartite(int[][] graph) {
        int n = graph.length;
        visited = new boolean[n];
        color = new boolean[n];
        for(int i = 0; i < n; i ++){
            if(!visited[i]){
                traverse(graph, i);
            }
        }
        return ok;
    }
    public void traverse(int[][] graph, int v){
        if(!ok){
            return;
        }
        visited[v] = true;
        for(int e : graph[v]){
            if(!visited[e]){
                color[e] = !color[v];
                traverse(graph,e);
            }else{
                if(color[e] == color[v]){
                    ok = false;
                    return;
                }
            }
        }
    }
}

二、力扣886. 可能的二分法

class Solution {
    boolean ok = true;
    boolean[] visited;
    boolean[] color;
    public boolean possibleBipartition(int n, int[][] dislikes) {
        visited = new boolean[n];
        color = new boolean[n];
        List<Integer>[] graph = builderGraph(dislikes, n);
        for(int i = 0; i < n; i ++){
            if(!visited[i]){
                BFS(graph, i);
            }
        }
        return ok;
    }
    public List<Integer>[] builderGraph(int[][] dislikes,int n){
        List<Integer>[] graph = new LinkedList[n];
        for(int i = 0; i < n ; i ++){
            graph[i] = new LinkedList<>();
        }
        for(int[] arr : dislikes){
            int to = arr[0]-1;
            int from = arr[1]-1;
            graph[to].add(from);
            graph[from].add(to);
        }
        return graph;
    }
    public void BFS(List<Integer>[] graph, int v){
        if(!ok){
            return;
        }
        Deque<Integer> deq = new LinkedList<>();
        deq.offerLast(v);
        visited[v] = true;
        while(!deq.isEmpty()){
            int cur = deq.pollFirst();
            for(int e : graph[cur]){
                if(!visited[e]){
                    visited[e] = true;
                    color[e] = !color[cur];
                    deq.offerLast(e);
                }else{
                    if(color[e] == color[cur]){
                        ok = false;
                        return;
                    }
                }
            }
        }
    }
}