[leetcode 277]Find the Celebrity---------找名人

Question:

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.

分析：

找名人，有n个人，这n个人中可能有一个名人，也可能没有名人。如果有名人，则其余的n-1个人都认识他，而这个名人谁都不认识。

可以通过问A,B是否认识来判别和排除候选人。

如果A认识B，则排除了A，因为名人不认识任何人；如果A不认识B，则排除了B，因为名人都被人认识。最初候选人为所有人。如果没有名人返回-1，有返回其编号

代码如下（参考网上大神）：

<span style="font-size:14px;">// Forward declaration of the knows API.
bool knows(int a, int b);

class Solution {
public:
    int findCelebrity(int n) {
        vector<int> candidates(n);
        iota(candidates.begin(), candidates.end(), 0);
        while (candidates.size() >= 2) {
            int a = candidates.back(); candidates.pop_back();
            int b = candidates.back(); candidates.pop_back();
            if (knows(a, b)) candidates.push_back(b);
            else candidates.push_back(a);
        }
        return verifyCelebrity(candidates[0], n);
    }
private:
    int verifyCelebrity(int c, int n) {
        for (int i = 0; i < n; i++) {
            if (i == c) continue;
            if (knows(c, i) || !knows(i, c))
                return -1;
        }
        return c;
    }
};</span>

简单版本：

<span style="font-size:14px;">// Forward declaration of the knows API.
bool knows(int a, int b);

class Solution {
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) continue;
            if (!knows(i, c) || knows(c, i))
                return -1;
        }
        return c;
    }
};</span>