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.

这题提示尽量少的调用knows函数来判断是否存在其他人都认识他，但是他不认识其他人的名人。

这题可以从图的遍历来讲，找那个入度为n-1,出度为0的结点，但是时间复杂度是O(n^2),边的数目最差是是O(n^2)级别。空间复杂度为O(n)。

该解法可以优化，用一些策略：

主要是先找candidate,之后验证这个candidate是否存在。

详细解释参见：

leetcode.com/problems/find-the-celebrity/discuss/71268/Straight-forward-C++-solution-with-explaination/157109

时间复杂度O(n),空间复杂度O(1).

代码：

# The knows API is already defined for you.
# @param a, person a
# @param b, person b
# @return a boolean, whether a knows b
# def knows(a, b):

class Solution(object):
    def findCelebrity(self, n):
        """
        :type n: int
        :rtype: int
        """
        candidate = 0
        #这一边扫描也可以利用two pointer来做：http://buttercola.blogspot.com/2015/09/leetcode-find-celebrity.html
        for i in xrange(1, n):
            if not knows(i, candidate):
                candidate = i
                
        for i in xrange(n):
            if i != candidate:
                if not knows(i, candidate) or knows(candidate, i):
                    return -1
        
        return candidate