Find the Celebrity 解答

Question

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 - 1people 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.

Solution

这题的关键是排除法。A知道B，那么我们就可以知道A一定不是candidate。所以我们就看B知道谁。

两层循环。

第一层找到candidate。

因为有且仅有一个celebrity，而且所有人都知道celebrity。所以如果有celebrity，通过这层循环一定可以找到它。

第二层验证candidate是否符合celebrity条件。

/* 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;
        // Exclusion way to find candidate
        for (int i = 1; i < n; i++) {
            if (knows(candidate, i)) {
                candidate = i;
            }
        }
        // Check whether candidate is valid
        for (int i = 0; i < n; i++) {
            if (i == candidate) {
                continue;
            }
            if (!knows(i, candidate) || knows(candidate, i)) {
                return -1;
            }
        }
        return candidate;
    }
}