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.

Solution

遍歷每一個(gè)人，看其余的n-1個(gè)人是否都認(rèn)識(shí)這個(gè)人，如果是，那么如果這個(gè)人不認(rèn)識(shí)任何人則輸出，否則繼續(xù)下一個(gè)人。

// Forward declaration of the knows API.
bool knows(int a, int b);

class Solution {
public:
    int findCelebrity(int n) {
        for(int i = 0; i < n; i++) {
            bool flag = false;
            for(int j = 0; j < n; j++) {
                if (i != j && !knows(j, i)) {
                    flag = true;
                    break;
                }
            }
            if (!flag) {
                flag = false;
                for(int j = 0; j < n; j++) {
                    if (i != j && knows(i, j)) {
                        flag = true;
                        break;
                    }
                }
                if (!flag) {
                    return i;
                }
            }
        }
        return -1;
    }
};

優(yōu)化

O(n).

以下是第一重循環(huán)的解釋。

Let's say i = k, you first time switch candidate, then 0 ~ k -1 will not have candidate. 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. So if the pre candidate(0) doesn't know 1 ~ k-1, then 1 ~ k - 1 are not celebrity. As you just switch candidate, then 0 is not.

code

// Forward declaration of the knows API.
bool knows(int a, int b);

class Solution {
public:
    int findCelebrity(int n) {
        int candidate = 0;
        // If candidate does not know k (0 ~ i-1). It means that k cannot be celebrity, since celebrity
        // must be known by everyone.
        for(int i = 1; i < n; i++) {
            if (knows(candidate, i)) {
                candidate = i;
            }
        }
    
        for(int j = 0; j < n; j++) {
            if (candidate != j && (knows(candidate, j) || !knows(j, candidate))) {
                return -1;
            }
        }
    
        return candidate;
    }
};