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.

题意：

有一坨人，只允许你问“谁认识谁”。得想个法儿把名人给找出来。什么叫名人？谁都认识TA，TA谁也不认。

想起来牛姐有木有？

Solution：

只要call一次 knows(i, j) by comparing a pair (i, j),  we are able to discard one of them

1.Find a candidate by one pass: make sure other people are not celebrities.if knows(i, j) is true,  i is guaranteed not to be celebrityotherwise, j is not a celebrity

2.Make sure if the candidate is the celebrity by one pass

code

 /* 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 candidate = 0;

     // assume candidate is celebrity

     // if candidate knows i, such candidate isn't celebrity, try another one

     for(int i = 1; i < n; i++){

         if(knows(candidate, i))

             candidate = i;

     }

     for(int i = 0; i < n; i++){

         // we don't want to check if person knows himself

         // if candidate knows i, such candidate isn't celebrity

         // if nobody knows candidate, such candidate isn't celebrity

         if(candidate!=i && (knows(candidate, i) || !knows(i, candidate))) return -1;

     }

     return candidate;

     }

 }