Leetcoed 277 Find the Celebrity

Find the Celebrity

Total Accepted: 1126 Total Submissions: 3603 Difficulty: Medium

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

.

转自
http://blog.csdn.net/xudli/article/details/48749295
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

.

[思路]

当 a -> b时, 可以推出,  a不可能是celebrity, b被人知道的数目+1... 用bitmap记录. 

/* The knows API is defined in the parent class Relation.
boolean knows(int a, int b); */

public class Solution extends Relation {
public int findCelebrity(int n) {
int[] bitmap = new int
;

for(int i=0; i<n; i++) {
for(int j=0; j<n; j++) {
if(i==j) continue;

if(bitmap[j]>=0) {
if( knows(i, j) ) {
bitmap[i] = -1;
bitmap[j]++;
} else {
bitmap[j] = -1;
}
}
}
}

for(int i=0; i<n; i++) {
if(bitmap[i] == n-1) {
for(int j=0; j<n; j++) {
if(i==j) continue;
if(knows(i, j)) return -1;
}
return i;
}
}

return -1;
}
}