Find the Celebrity

https://leetcode.com/problems/find-the-celebrity/description/

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.

Thoughts

knows(a, b) = true意味着a认识b, a肯定不是名人, b可能是名人, a排除; knows(a,b) = false, 意味着b肯定不是名人, 因为a不认识它, b排除. 因此只要一遍遍历, 每次都能排除一个元素, 剩下的成为候选. 但这么做只能保证排除了的肯定不是名人, 但不能保证最后剩下的candidate一定是名人, 因此还要一次遍历进行验证.

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 res = 0;
        for (int i = 1; i < n; i++) {
            res = knows(res, i) ? i : res;
        }

        for (int i = 0; i < n; i++) {
            if (i != res && (!knows(i, res) || knows(res, i))) {
                return -1;
            }
        }

        return res;
    }
}

Analysis

时间复杂度O(N).