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.

/* The knows API is defined in the parent class Relation.
boolean knows(int a, int b); */ public class Solution extends Relation {
public int findCelebrity(int n) {
int res = 0;
// find the celebrity
for (int i = 0; i < n; i++) {
if (knows(res, i)) {
res = i;
}
} // return -1 if res is not celebrity
for (int i = 0; i < n; i++) {
if (i != res && (knows(res, i) || !knows(i, res))) {
return -1;
}
}
return res;
}
}

[LC] 277. Find the Celebrity的更多相关文章

  1. 名人问题 算法解析与Python 实现 O(n) 复杂度 (以Leetcode 277. Find the Celebrity为例)

    1. 题目描述 Problem Description Leetcode 277. Find the Celebrity Suppose you are at a party with n peopl ...

  2. [LeetCode] 277. Find the Celebrity 寻找名人

    Suppose you are at a party with n people (labeled from 0 to n - 1) and among them, there may exist o ...

  3. 277. Find the Celebrity

    题目: Suppose you are at a party with n people (labeled from 0 to n - 1) and among them, there may exi ...

  4. [LeetCode#277] Find the Celebrity

    Problem: Suppose you are at a party with n people (labeled from 0 to n - 1) and among them, there ma ...

  5. LeetCode 277. Find the Celebrity (找到明星)$

    Suppose you are at a party with n people (labeled from 0 to n - 1) and among them, there may exist o ...

  6. [leetcode]277. Find the Celebrity 找名人

    Suppose you are at a party with n people (labeled from 0 to n - 1) and among them, there may exist o ...

  7. 【LeetCode】277. Find the Celebrity 解题报告 (C++)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 暴力 日期 题目地址:https://leetcode ...

  8. [leetcode]277. Find the Celebrity谁是名人

    Suppose you are at a party with n people (labeled from 0 to n - 1) and among them, there may exist o ...

  9. <Array> 277 243 244 245

    277. Find the Celebrity knows(i, j): By comparing a pair(i, j), we are able to discard one of them 1 ...

随机推荐

  1. Necroptosis|Apoptosis|CTC|

    大数据-外周血研究 Necroptosis与Apoptosis...区别:肿瘤导致和自身凋亡. CTC,肿瘤细胞脱落进入外周血,CTC(循环肿瘤细胞,CirculatingTumorCell)是存在于 ...

  2. ZJNU 2353 - UNO

    大模拟,但是题目好像有些地方表述不清 根据UNO在初中曾被别人虐了很久很久的经历 猜测出了原本的题意 本题中的+2虽然有颜色,但是也可以当作原UNO游戏中的+4黑牌 即在某人出了+2后,可以出不同颜色 ...

  3. Java if、switch语句,break,case,类型转换、常量、赋值比较、标识符(2)

    if语句: /* if else 结构 简写格式: 变量 = (条件表达式)?表达式1:表达式2: 三元运算符: 好处:可以简化if else代码. 弊端:因为是一个运算符,所以运算完必须要有一个结果 ...

  4. Python笔记_第一篇_面向过程_第一部分_2.内存详解

    Python的很多教材中并没有讲内存方面的知识,但是内存的知识非常重要,对于计算机工作原理和方便理解编程语言是非常重要的,尤其是小白,因此需要把这一方面加上,能够更加深入的理解编程语言.这里引用了C语 ...

  5. ios 真机使用相机闪退问题

    需要增加权限 在info文件增加 主要前面的key 后面的cameraDesciption只是一段描述 随便写就可以

  6. ACM-ICPC Asia Beijing Regional Contest 2018 Reproduction hihocoder1870~1879

    ACM-ICPC Asia Beijing Regional Contest 2018 Reproduction hihocoder1870~1879 A 签到,dfs 或者 floyd 都行. #i ...

  7. httpsqs 源码修改(内部自动复制多队列)

    /* HTTP Simple Queue Service - httpsqs v1.7 Author: Zhang Yan (http://blog.s135.com), E-mail: net@s1 ...

  8. python文件读写 文件修改

    #设置一个变量f为文件对象,并打开文件#写文件#f = open('user.txt','w',encoding='utf-8') #f是一个文件对象f=open(r'c:\Users\PL\Desk ...

  9. ruoyi IpUtils

    package com.ruoyi.common.utils; import java.net.InetAddress; import java.net.UnknownHostException; i ...

  10. Perl:正则中问号的四周用途:1.字面意义的问号 2. 量词 3. 表示非贪心的修饰符 4.用以表示不具有记忆功能的圆括号

    Perl:正则中问号的四周用途:1.字面意义的问号  2. 量词   3. 表示非贪心的修饰符  4.用以表示不具有记忆功能的圆括号 非贪心:在量词后面加?即可