277. Find the Celebrity


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.


/* 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;
//每一次比較只有兩種情況,knows(a, b)是true的話,那麼a肯定不是candidate; knows(a, b)是false的話,那麼b肯定不是candidate. 所以一直比較到最後會留下一個candidate,然後我們再驗證這個是不是正解。
for (int i = 1; i < n; i  ) {
if (knows(candidate, i)) {
candidate = i;
for (int i = 0; i < n; i  ) {
if (candidate != i && (knows(candidate, i) || !knows(i, candidate))) {
return -1;
return candidate;