277. Find the Celebrity

Problem:

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.

Analysis:

two pass,第一次扫描得到candidate,第二次扫描验证candidate. 在一次扫描的时候，如果knows(c, i) 说明c不是candidate，更新candidate到i. 如果!knows(c, i), 说明i不是candidate，继续向后走。

Solution:

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public class Solution extends Relation { public int findCelebrity(int n) { int c = 0; for (int i = 1; i < n; i++) { if (knows(c, i)) c = i; } for (int i = 0; i < n; i++) { if (i != c && (knows(c, i) || !knows(i, c))) return -1; } return c; } }