zoukankan      html  css  js  c++  java
  • LeetCode 886. Possible Bipartition

    原题链接在这里:https://leetcode.com/problems/possible-bipartition/

    题目:

    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].

    题解:

    To determine it is bipartition, we could see if we could color them into 2 different colors.

    First use dislikes to construct a graph.

    For the node hasn't been color before and it is in the graph, put it as one color 1, then perform BFS.

    For cur polled number, if its neighbor has the same color, then return false.

    Time Complexity: O(N+E). E = dislikes.length.

    Space: O(N).

    AC Java:

     1 class Solution {
     2     public boolean possibleBipartition(int N, int[][] dislikes) {
     3         Map<Integer, Set<Integer>> graph = new HashMap<>();
     4         for(int [] d : dislikes){
     5             graph.putIfAbsent(d[0], new HashSet<Integer>());
     6             graph.putIfAbsent(d[1], new HashSet<Integer>());
     7             
     8             graph.get(d[0]).add(d[1]);
     9             graph.get(d[1]).add(d[0]);
    10         }
    11         
    12         int [] color = new int[N+1];
    13         for(int i = 1; i<=N; i++){
    14             if(color[i]==0 && graph.containsKey(i)){
    15                 color[i] = 1;
    16                 LinkedList<Integer> que = new LinkedList<>();
    17                 que.add(i);
    18                 while(!que.isEmpty()){
    19                     int cur = que.poll();
    20                     for(int nei : graph.get(cur)){
    21                         if(color[nei] == 0){
    22                             color[nei] = -color[cur];
    23                             que.add(nei);
    24                         }else if(color[nei] == color[cur]){
    25                             return false;
    26                         }
    27                     }
    28                 }
    29             }
    30         }
    31         
    32         return true;
    33     }
    34 }

    类似Is Graph Bipartite?.

  • 相关阅读:
    springBoot单元测试-模拟MVC测试
    springBoot单元测试-基础单元测试
    java使用HttpClient 发送get、pot请求
    定时任务框架-quartz 时间配置
    定时任务框架-quartz
    java接入极光推送
    实现自动解析properties文件并装配到Bean
    Redis-NoSql 概述,NoSql的优点
    springboot
    JAVA常见集合类
  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/11949381.html
Copyright © 2011-2022 走看看