Find the Town Judge

In a town, there are N people labelled from 1 to N.  There is a rumor that one of these people is secretly the town judge.

If the town judge exists, then:

1. The town judge trusts nobody.
2. Everybody (except for the town judge) trusts the town judge.
3. There is exactly one person that satisfies properties 1 and 2.

You are given trust, an array of pairs trust[i] = [a, b] representing that the person labelled a trusts the person labelled b.

If the town judge exists and can be identified, return the label of the town judge.  Otherwise, return -1.

Example 1:

Input: N = 2, trust = [[1,2]]
Output: 2

Example 2:

Input: N = 3, trust = [[1,3],[2,3]]
Output: 3

Example 3:

Input: N = 3, trust = [[1,3],[2,3],[3,1]]
Output: -1

Example 4:

Input: N = 3, trust = [[1,2],[2,3]]
Output: -1

Example 5:

Input: N = 4, trust = [[1,3],[1,4],[2,3],[2,4],[4,3]]
Output: 3

Note:

1. 1 <= N <= 1000
2. trust.length <= 10000
3. trust[i] are all different
4. trust[i][0] != trust[i][1]
5. 1 <= trust[i][0], trust[i][1] <= N

题目理解：

给定一组人与人之间的信任关系，[a,b]表示a信任b。定义judge：judge这个人不相信任何人，但是被所有人相信。在给定的信任关系中，找出judge，如果不能找出，返回-1

解题思路：

记录所有人信任的数目和被人信任的数目，如果有一个人的信任数是0，而被信任数是N-1，那么他就是judge。如果没有这样的人，则找不出judge

class Solution {
    public int findJudge(int N, int[][] trust) {
        int[][] record = new int[N + 1][2];
        for(int[] it : trust){
            record[it[0]][0]++;
            record[it[1]][1]++;
        }
        for(int i = 1; i < N + 1; i++)
            if(record[i][0] == 0 && record[i][1] == N - 1)
                return i;
        return -1;
    }
}