zoukankan      html  css  js  c++  java
  • [leetcode]785. Is Graph Bipartite? [bai'pɑrtait] 判断二分图

    Given an undirected graph, return true if and only if it is bipartite.

    Example 1:
    Input: [[1,3], [0,2], [1,3], [0,2]]
    Output: true
    Explanation: 
    The graph looks like this:
    0----1
    |    |
    |    |
    3----2
    We can divide the vertices into two groups: {0, 2} and {1, 3}.

    Example 2:
    Input: [[1,2,3], [0,2], [0,1,3], [0,2]]
    Output: false
    Explanation: 
    The graph looks like this:
    0----1
    |   |
    |   |
    3----2
    We cannot find a way to divide the set of nodes into two independent subsets.

    设G=(V,E)是一个无向图。如顶点集V可分割为两个互不相交的子集V1,V2之并,并且图中每条边依附的两个顶点都分别属于这两个不同的子集

    思路

    1. based on Graph Bipartite attribute, we can fill two different color for each subset.
    2. if not Graph Bipartite, at lease one node such that its color happens to be the same as its neighbor
    3. coz we need to traversal each node, we can both use dfs and bfs

    代码

     1 class Solution {
     2     public boolean isBipartite(int[][] graph) {
     3         int[] visited = new int[graph.length];
     4         //default 0: not visited;
     5         //lable 1: green
     6         //lable 2: red     
     7         for(int i = 0; i < graph.length; i++) {
     8             // such node has been visited
     9             if(visited[i] != 0) {continue;}    
    10             //such node has not been visited
    11             Queue<Integer> queue = new LinkedList();
    12             queue.add(i);
    13             // mark as green
    14             visited[i] = 1;
    15             while(!queue.isEmpty()) {
    16                 int cur = queue.poll();
    17                 int curLable = visited[cur];
    18                 // if curLable is green, fill neighborLable to red
    19                 int neighborLable = curLable == 1? 2:1;
    20                 for(int neighbor:graph[cur]) {
    21                     //such node has not been visited
    22                     if(visited[neighbor] == 0) {
    23                         visited[neighbor] = neighborLable;
    24                         queue.add(neighbor);
    25                     }  
    26                     // node visited, and visited[neighbor] != neighborLable, conflict happens
    27                     else if(visited[neighbor] != neighborLable) {
    28                         return false;
    29                     }
    30                 }
    31             }       
    32         }  
    33         return true;
    34     }
    35 }
  • 相关阅读:
    openresty 使用 log_by_lua 发送日志到 syslog-ng
    uuid 了解
    基于openresty 的几个开发框架
    openresty 几个插件使用
    kong 了解
    openresty && hashids&& redis 生成短链接
    kong k8s 安装 以及可视化管理界面
    hashids 了解
    Apache Tez 了解
    Cascalog了解
  • 原文地址:https://www.cnblogs.com/liuliu5151/p/9814350.html
Copyright © 2011-2022 走看看