原題
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.
解題思路
- 第一個(gè)for循環(huán)价匠，從第0個(gè)人開始冕杠，如果k是第0個(gè)人認(rèn)識(shí)的第一個(gè)人，說明1到k-1這些人0不認(rèn)識(shí)逗宜，所以排除了名人的可能。按照此規(guī)則進(jìn)行下去霹肝，最后candidate停在某一個(gè)位置，這個(gè)位置后面一定也沒有名人，因?yàn)橛械脑捴黄茫琧andidate會(huì)update等于它
- 最后檢查candidate對(duì)不對(duì)
完整代碼
# 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
