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:

The town judge trusts nobody.
Everybody (except for the town judge) trusts the town judge.
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
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
 

Constraints:

1 <= n <= 1000

0 <= trust.length <= 104
trust[i].length == 2
All the pairs of trust are unique.
ai != bi
1 <= ai, bi <= n

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/find-the-town-judge
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这题要看成一个有向图，法官的出度为0，入度不止为一，且法官唯一。我们需要做的就是信任的时候加一：

 

class Solution {
   public int findJudge(int n, int[][] trust) {
      int[] trustValue = new int[n+1];//数组初始值为0
      
      for(int[] i : trust){
         trustValue[i[0]]--;
         trustValue[i[1]]++;
      }
   
      for(int i = 1;i <= n;i++){
         if(trustValue[i] == (n-1)) return n;
      }
      return -1;
   }
}