Trivia leetcode-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.
原题地址

比较蠢的办法就是对每个人查看，如果他不认识其他人，并且其他人认识他，那他就是celebrity。Time: O(n²).

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

更好的办法是Two pass，第一遍先找到可能是celebrity的人，第二遍验证。Time: O(n).

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bool (int a, int b);
class Solution {
public:
    int findCelebrity(int n) {
        int cel = 0;
        // if a knows b, a must not be the celebrity
        // if b is the celebrity, a will know him, so we won't miss him
        for (int i = 1; i < n; i++) {
            if (knows(cel, i))
                cel = i;
        }
        // validate
        for (int i = 0; i < n; i++) {
            if (i != cel && (knows(cel, i) || !knows(i, cel)))
                return -1;
        }
        return cel;
    }
};