997. Find the Town Judge-找到小镇的法官

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

Constraints:
1 <= N <= 1000
0 <= trust.length <= 10^4
trust[i].length == 2
trust[i] are all different
trust[i][0] != trust[i][1]
1 <= trust[i][0], trust[i][1] <= N

Solution in python:

class Solution:
    def findJudge(self, N: int, trust: List[List[int]]) -> int:
        adic = dict()
        aset = set()
        K = -1
        for item in trust:
            aset.add(item[0])
            if item[1] not in adic.keys():
                adic[item[1]] = 1
            else:
                adic[item[1]] += 1
        for item in adic.items():
            print(item[0], item[1])
            if item[1] == N-1 and (item[0] not in aset):
                K = item[0]
        if N == 1 and not adic:
            K = 1
        return K