LeetCode997.Find the Town Judge（找到小镇的法官）

Description

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.

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在一个小镇里，按从 1 到 N 标记了 N 个人。传言称，这些人中有一个是小镇上的秘密法官。

如果小镇的法官真的存在，那么：

1. 小镇的法官不相信任何人。

2. 每个人（除了小镇法官外）都信任小镇的法官。

3. 只有一个人同时满足属性 1 和属性 2 。

给定数组 trust，该数组由信任对 trust[i] = [a, b] 组成，表示标记为 a 的人信任标记为 b 的人。

如果小镇存在秘密法官并且可以确定他的身份，请返回该法官的标记。否则，返回 -1。

题目链接：https://leetcode.com/problems/find-the-town-judge/
个人主页：http://redtongue.cn or https://redtongue.github.io/

Difficulty: easy

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 <= N <= 1000

• trust.length <= 10000

• trust[i] are all different

• trust[i][0] != trust[i][1]

• 1 <= trust[i][0], trust[i][1] <= N

分析

• 初始化字典A，每一项代表每个人被那些人信任；

• 对于任意trust，a信任b，从A中删除a；

• 同时若b在A中，则将a添加为信任b的list中；

• 返回A中被信任N-1次的人，若无，返回-1。

参考代码

class Solution:
def findJudge(self, N: int, trust: List[List[int]]) -> int:
    A={}
    for i in range(N):
        A[i+1]=[]
    for t in trust:
        x,y=t
        if(x in A):
            A.pop(x)
        if(y in A):
            A[y].append(x)
    for a in A:
        if(len(A[a])==N-1):
            return a
    return -1