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

    package LeetCode_886
    
    import java.util.*
    import kotlin.collections.ArrayList
    
    /**
     * 886. Possible Bipartition
     * https://leetcode.com/problems/possible-bipartition/description/
     *
     * 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]
    
    Constraints:
    1. 1 <= N <= 2000
    2. 0 <= dislikes.length <= 10000
    3. dislikes[i].length == 2
    4. 1 <= dislikes[i][j] <= N
    5. dislikes[i][0] < dislikes[i][1]
    6. There does not exist i != j for which dislikes[i] == dislikes[j].
     * */
    class Solution {
        /*
        * solution: Create graph + BFS, Time complexity:O(V+E), Space complexity:O(V+E)
        * */
        fun possibleBipartition(N: Int, dislikes: Array<IntArray>): Boolean {
            val graph = Array(N + 1, { ArrayList<Int>() })
            for (i in 0 until N) {
                graph[i] = ArrayList()
            }
            //create graph
            for (dislike in dislikes) {
                graph[dislike[0]].add(dislike[1])
                graph[dislike[1]].add(dislike[0])
            }
    
            //-1:non-color, 0:red, 1:blue
            val corlorArray = IntArray(N + 1, { -1 })
            for (i in 0 until N) {
                if (corlorArray[i] != -1) {
                    continue
                }
                corlorArray[i] = 0
                val queue = LinkedList<Int>()
                queue.offer(i)
                while (queue.isNotEmpty()) {
                    val cur = queue.pop()
                    for (adj in graph[cur]!!) {
                        if (corlorArray[cur] == corlorArray[adj]) {
                            return false
                        } else {
                            //give opposite color to neighbor
                            if (corlorArray[adj] == -1) {
                                corlorArray[adj] = corlorArray[cur].xor(1)
                                queue.offer(adj)
                            }
                        }
                    }
                }
            }
            return true
        }
    }

     Similar problem:  785. Is Graph Bipartite?

  • 相关阅读:
    SQL SERVER 博客
    deadlock
    dbcc 官方文档
    HOW TO CHECK FOR ACTIVE TRACE FLAGS ON MICROSOFT SQL SERVER
    Enable a SQL Server Trace Flag Globally on Linux
    SQL server 2008 安装问题解决 转
    Looking deeper into SQL Server using Minidumps
    sql-server-linux 官网
    破解SQLServer for Linux预览版的3.5GB内存限制 (RHEL篇) 转
    SQL Server on Linux: How? Introduction: SQL Server Blog
  • 原文地址:https://www.cnblogs.com/johnnyzhao/p/13493977.html
Copyright © 2011-2022 走看看