In a town, there are n people labeled 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 an array trust where trust[i] = [ai, bi] representing that the person labeled ai trusts the person labeled bi.

Return the label of the town judge if the town judge exists and can be identified, or return -1 otherwise.

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

     求出town judge 有两点要求，judge 不相信任何人，任何其他人都相信judge（注意其他人可以除了judge以外还相信其他人，而不仅仅只有一个judge）。也就是judge一定被n-1个人相信。这道题可以利用图论解决，trust数组其实就是各个结点的临接表，如果a相信b（a->b），则b的入度加一，a的出度加一（等价于入度减一），利用一个数组维护所有人（结点）的入度关系，然后寻找入度为n-1的结点即为judge。

    

class Solution {
public:
    int findJudge(int n, vector<vector<int>>& trust) {
       vector<int> graph(n+1,0);
       for(auto tt:trust){
           graph[tt[0]]--;
           graph[tt[1]]++;
       } 
        for(int i=1;i<n+1;++i){
            if(graph[i]==n-1){
                return i;
            }
        }
        return -1;
        
    }
};