Find the Celebrity

题目地址：
https://leetcode.com/problems/find-the-celebrity/#/description

题目：

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.

解题思路：
这道题首先要找出一个potential的rst，然后利用给的api来实时更新rst，最后再loop一遍判断。这里会有一个变种题，就是如果>,<没有传递性，即A>B, B > C 不能推出A>C，那么要找出array里面最大的数。

代码：

public int findCelebrity(int n) {
    int rst = 0;
    for(int i = 1; i < n; i++){
        // rst is guranteed to be updated and won't change if there is celebrity        
        if(knows(rst, i)){
            rst = i;
        }
    }
    for(int i = 0; i < n; i++){
        if(i != rst && (knows(rst, i)) || !knows(i, rst)){
            return -1;
        }
    }
    return rst;
}

变种：

public static void main(String[] args){
    FindtheCelebrity findtheCelebrity = new FindtheCelebrity();
    int[] nums = {3, 4, 1, 8, 5, 2, 9, 6};
    int rst = findtheCelebrity.findCelebrity(nums);
    System.out.println(rst);
}
public static int findCelebrity(int[] nums) {
    if(nums == null || nums.length == 0){
        return -1;
    }
    List<Integer> less = new ArrayList<>();
    int rst = nums[0];
    int len = nums.length;
    for(int i = 1; i <= len - 1; i++){
        if(rst < nums[i]){
            less.add(rst);
            rst = nums[i];
        }
    }
    for(int i = 0; i <= less.size() - 1; i++){
        if(rst < less.get(i)){
            return -1;
        }
    }
    return rst;
}