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.

 

two pass

先遍历一次数组，找出可能的候选，p从0开始，如果p认识i，那么p不符合条件，i符合条件，把 i 赋值给p

再遍历一次数组，如果当前值 i 不为p，并且如果i 不认识p或者p认识i，说明之前找到的p不符合条件，返回-1

如果第二次遍历结束没有不符合条件，返回p

time: O(N), space: O(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 p = 0;        for(int i = 1; i < n; i++) {            if(knows(p, i))                p = i;        }                for(int i = 0; i < n; i++) {            if(i != p && (!knows(i, p) || knows(p, i)))                return -1;        }        return p;    }}

 

 

reference: https://leetcode.com/problems/find-the-celebrity/discuss/71227/Java-Solution.-Two-Pass