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  • 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?

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  • 原文地址:https://www.cnblogs.com/johnnyzhao/p/13493977.html
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