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# 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`.

## Example

**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`.

## Note

模拟一下这个过程：

第一轮遍历：如果a认识b，b就可能是目标，遍历所有人得到我们的目标

第二轮去验证：要求目标不认识任何人且任何人认识目标，违反就输出-1

## Code

```java
/* 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 res = 0;
        for (int i = 0; i < n; i++) {
            if (knows(res, i)) {
                res = i;
            }
        }
        for (int i = 0; i < n; i++) {
            if (i != res) {
                if (knows(res, i) || !knows(i, res)) {
                    return -1;
                }
            }
        }
        return res;
    }
}
```