### 原题

Suppose you are at a party with n people (labeled from 0 ton - 1) and among them, there may exist one celebrity. The definition of a celebrity is that all the othern - 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 functionint findCelebrity(n), your function should minimize the number of calls toknows.

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.

### 解题思路

- 第一个for循环，从第0个人开始，如果k是第0个人认识的第一个人，说明1到k-1这些人0不认识，所以排除了名人的可能。按照此规则进行下去，最后candidate停在某一个位置，这个位置后面一定也没有名人，因为有的话，candidate会update等于它
- 最后检查candidate对不对

### 完整代码

```
# 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):
pass
class Solution(object):
def findCelebrity(self, n):
"""
:type n: int
:rtype: int
"""
candidate = 0
for i in range(1, n):
if knows(candidate, i):
candidate = i
for i in range(n):
if candidate != i and (knows(candidate, i) or knows(i, candidate)):
return -1
return candidate
```