Find celebrity

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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.

分析：

我们从0开始，依次问他是否认识他的下一个，比如 1， 如果 0 说认识，那么 0 一定不是名人，1 有可能是名人； 如果0 说不认识，1 一定不是名人，0 却有可能是名人。这是这个方法巧的地方。所以，不管1 回答是还是不是，我们都可以确定其中一个不是名人，然后，我们继续问 可能是名人那一位 是否认识 2， 然后依次下去，直到第 N - 1个人。这样我们就可以只要问 （N - 1） 个问题就可以把名人找出来。

 1     int findCelebrity(int n) {
 2         int c = 0;
 3         for (int i = 1; i < n; ++i) {
 4             if (knows(c, i)) {
 5                 c = i;
 6             }
 7         }
 8 
 9         for (int i = 0; i < n; ++i) {
10             if (c != i && (knows(c, i) || !knows(i, c)))
11                 return -1;
12         }
13         return c;
14     }

Find celebrity

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原文地址：http://www.cnblogs.com/beiyeqingteng/p/5743580.html