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.
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這題看着特別複雜，其實邏輯超級簡單。有n個人，編號從0到n-1.有一個或者0個celebrity。我們可以traverse 每個人兩遍就可以找到這個celebrity了。
條件：every one knows x && x doesn’t known anyone.
所以O(n) code:
class Solution {
public:
int findCelebrity(int n) {
int candidate = 0;
for(int i = 1; i < n; ++i){
if(knows(candidate,i)) candidate = i;// or !(i,candidate) : current people dont know candidate;
}
for(int i = 0; i<n; ++i){
if( i != candidate && (!knows(i, candidate) || knows(candidate, i))) return -1;
}
return candidate;
}
};之所以可以這樣做，是因爲celebrity number <= 1!
