标签：ini 题意 eal help i++ orm cti xtend 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.
考察题意的理解和转化, 找到knows(候选人, i) 来传递候选人, 候选人谁也不知道.
最后确定是否是候选人, 因为前面的可能有人不知道他, 那么他就不是候选人
最佳做法：O（N）time， O(1)space
The first pass is to pick out the candidate. If candidate knows i, then switch candidate. The second pass is to check whether the candidate is real.
public class Solution extends Relation {
public int findCelebrity(int n) {
int candidate = 0;
for(int i = 1; i < n; i++){
if(knows(candidate, i))
candidate = i;
}
for(int i = 0; i < n; i++){
if(i != candidate && (knows(candidate, i) || !knows(i, candidate))) return -1;
}
return candidate;
}
}
标签：ini 题意 eal help i++ orm cti xtend question
原文地址：http://www.cnblogs.com/apanda009/p/7357234.html
