Leetcode 277: Find the Celebrity

最新推荐文章于 2019-03-15 14:08:54 发布

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最新推荐文章于 2019-03-15 14:08:54 发布

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原文链接：http://www.cnblogs.com/liangmou/p/8369650.html

本文介绍了一种高效算法，用于从n个人中找出唯一的名人。名人定义为被其他所有人认识但自己不认识任何人的人。通过最少次数地调用knows(a, b)函数来判断a是否认识b，从而找到名人。

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), your function should minimize the number of calls to knows.

Note: 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.

 1 /* The Knows API is defined in the parent class Relation.
 2       bool Knows(int a, int b); */
 3 
 4 public class Solution : Relation {
 5     public int FindCelebrity(int n) {
 6         int candidate = 0;
 7         
 8         for (int i = 1; i < n; i++)
 9         {
10             // if this guy doesn't know the candidate or the candidate knows this guy, the candidate isn't the celebrity
11             // so we reset the candidate to this guy
12             if (!Knows(i, candidate) || Knows(candidate, i))
13             {
14                 candidate = i;
15             }
16         }
17         
18         for (int i = 0; i < n; i++)
19         {
20             if (i != candidate && (!Knows(i, candidate) || Knows(candidate, i)))
21             {
22                 return -1;
23             }
24         }
25         
26         return candidate;
27     }
28 }