Find the Celebrity

最新推荐文章于 2022-02-15 16:28:09 发布

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本文介绍了一个寻找特定人群中名人的算法问题。该问题的目标是最小化询问次数来确定是否存在一位被其他人所知但自己却不认识任何人的名人。文章提供了一种有效的解决方案，并详细解释了其工作原理。

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我认为这是一道智力题

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

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.

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