边工作边刷题：70天一遍leetcode: day 85

Find the Celebrity
要点：

• 这题从solution反过来想比较好：loop through n同时maintain一个candidate：如果cand认识某个i，那么modify cand。实际上检查另一个条件i不认识cand也可以，因为这2个条件是exclusive的
• 这个过程保证了cand不认识其之后的所有人。两个问题：
  • 一是会不会漏掉之前可能的celebrity呢？不会，如果当前cand不know i，i不可能是candidate。
  • 因为只是确保了之后的所有人，会不会是False positive呢？所以要再loop确认结果。所以两个loop一个验证必要条件，一个验证充分条件

https://repl.it/Cet1/1
错误点：

• 最后验证的时候要两个条件都验证，否则0 knows 1， 1 knows 0的情况会错

# 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.

# Hide Company Tags LinkedIn Facebook
# Hide Tags Array

# 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
        """
        cand = 0
        for i in xrange(1,n):
        	if knows(cand, i):
        		cand = i
        
        for i in xrange(n):
        	if i!=cand:
        		if not knows(i,cand) or knows(cand, i):
        			return -1
        return cand