标签：return knows people Celebrity there result 277 Find celebrity

package LeetCode_277

/**
 * 277. Find the Celebrity
 * (Prime)
 * Suppose you are at a party with n people (labeled from 0 to n - 1) and among them, there may exist one celebrity.
 * The definition of a celebrity is that all the other n - 1 people know him/her but he/she does not know any of them.
Now you want to find out who the celebrity is or verify that there is not one.
The only thing you are allowed to do is to ask questions like: "Hi, A. Do you know B?" to get information of whether A knows B.
You need to find out the celebrity (or verify there is not one) by asking as few questions as possible (in the asymptotic sense).
You are given a helper function bool knows(a, b) which tells you whether A knows B.
Implement a function int findCelebrity(n).
There will be exactly one celebrity if he/she is in the party.
Return the celebrity's label if there is a celebrity in the party.
If there is no celebrity, return -1.
 * */
class Solution {
    /*
    * solution: compare two times, Time complexity:O(n), Space complexity:O(1)
    * */
    fun findCelebrity(peoples: IntArray): Int {
        var result = 0
        //check the first time, fount out some one maybe a celebrity
        for (people in peoples) {
            //if a knows b (now result is a, people is b), so update people to result, result maybe a celerity
            //because just one people is celebrity
            if (knows(result, people)) {
                result = people
            }
        }

        for (people in peoples) {
            //if some one knows the people or people didn't him, he is not a celebrity
            if (people != result && (knows(result, people) || !knows(people, result))) {
                return -1
            }
        }

        return result
    }

    //whether a knows b
    private fun knows(a: Int, b: Int): Boolean {
        return false
    }
}

标签：return,knows,people,Celebrity,there,result,277,Find,celebrity
来源： https://www.cnblogs.com/johnnyzhao/p/13746867.html

本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享；
2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关；
3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关；
4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除；
5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。