997. Find the Town Judge

QUESTION:

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

EXPLANATION:

我的解法是：每个人都可以投给别人，那么就可以计算出每个人投给别人的票，和自己获得的票。

用一个二维数组来保存两个值。

当只有自己获得所有人信任，同时自己不信任任何其他人的时候，才能返回该值。

但是在细想之后，你会发现，如果这个人投票给了其他人，那么他就不能是法官。那么就可以节省很多判断。

SOLUTION:

class Solution {
    public int findJudge(int N, int[][] trust) {
        if(trust.length==0) return 1;
        int[][] result = new int[N+1][2];
        for(int i=0;i<trust.length;i++){
            result[trust[i][0]][0]++;//信任别人个数
            result[trust[i][1]][1]++;//被信任个数
        }
        for(int i = 0;i<result.length;i++){
            if(result[i][0]==0 && result[i][1]==N-1) return i;
        }
        return -1;
    }
}
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