zoukankan      html  css  js  c++  java
  • [LC] 277. Find the Celebrity

    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;
        }
    }
  • 相关阅读:
    字符串练习
    Python基础
    熟悉常见的Linux命令
    大数据概述
    实验三 递归下降分析法
    简化C语言文法
    实验一 词法分析程序实验
    词法分析程序
    制作首页的显示列表
    完成登录功能
  • 原文地址:https://www.cnblogs.com/xuanlu/p/11896128.html
Copyright © 2011-2022 走看看