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), 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.

Tips:

走两遍，O(n)。 第一遍先选候选人，candidate从0开始。如果candidate认识第i个人，说明当前人不是，第i个可能是。将candidate替换成i接着看他认不认识后面的人。

第二遍确认了他不认识大家，则开始检验是不是大家都认识他。

he first pass is to pick out the candidate. If candidate knows i, then switch candidate. The second pass is to check whether the candidate is real.

Code:

public class Solution extends Relation {
    public int findCelebrity(int n) {
        int candidate = 0;
        for(int i = 1; i < n; i++){
            if(knows(candidate, i))
                candidate = i;
        }
        for(int i = 0; i < n; i++){
            if(i != candidate && (knows(candidate, i) || !knows(i, candidate))) return -1;
        }
        return candidate;
    }
}