LeetCode 10 May, Find the Town Judge, 找出法官

Find the Town Judge

In a town, there are N people labbelled 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

分析：

如下图所示：假设3是法官，那么除了3自己，所有的人都相信3（如红色的1所示），并且3不相信任何人（如浅橙色的空白所示）。

最简单的想法就是，遍历一次列表的同时构建两个字典，一个字典记录trust[i][0]被信任的次数。一个字典记录trust[i][1]是否相信别人。然后遍历整个人员N，若该索引人被信任了N-1次，并且没在另一个字典中出现，那么这个人就是法官。否则返回-1.

Python3 代码如下：

# 方法一
# Runtime: 788ms
# Memory Usage: 18.4MB
class Solution:
    def findJudge(self, N: int, trust: List[List[int]]) -> int:
        if N == 1: return 1
        
        count = {}
        believe = {}
        for x in trust:
            if x[1] not in count:
                count[x[1]] = 1
            else:
                count[x[1]] += 1
            if x[0] not in believe:
                believe[x[0]] = 1
        for index in range(1, 1001):
            if index not in count: continue
            if (index not in believe) and (count[index] == N - 1): return index
        return -1