Leetcode: Find the Celebrity

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.

像一个擂台一样，一开始维护两个candidate：A 和 B，维护一个stack, 存储所有变量

if knows(A, B), A一定不是，B可能是，A换人：A = stack.pop()

if !knows(A, B), B一定不是，A可能是, B换人：B = stack.pop()

这个问题取巧在于know(A,B)无论如何会淘汰一个，经过N-1回合，一定只剩最后一个候选人AorB，假定是A

最后再验证一下他是不是就好了，

验证方法：所有除他之外的人E依次上擂，这次每个要比两回合

if knows(A,E), return -1

if !knows(E,A), return -1

/* 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) {
if (n == 1) return 0;
Stack<Integer> st1 = new Stack<Integer>();
for (int i=0; i<n; i++) {
st1.push(i);
}
Stack<Integer> st2 = (Stack<Integer>)st1.clone();
int A = st1.pop();
int B = st1.pop();
while (!st1.isEmpty()) {
if (knows(A, B)) {
A = st1.pop();
}
else {
B = st1.pop();
}
}
A = knows(A, B)? B : A; //only one candidate

//verify now;
while (!st2.isEmpty()) {
int V = st2.pop();
if (V != A) {
if (knows(A, V)) return -1;
if (!knows(V, A)) return -1;
}
}
return A;
}
}