[leetcode] 997. Find the Town Judge @ python

原题

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:

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 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 <= N <= 1000
trust.length <= 10000
trust[i] are all different
trust[i][0] != trust[i][1]
1 <= trust[i][0], trust[i][1] <= N

解法

graph. 先遍历trust, 构造graph字典, 然后遍历字典, 如果找到某个键的列表为空, 那么检查它是否符合法官的定义.

代码

class Solution(object):
    def findJudge(self, N, trust):
        """
        :type N: int
        :type trust: List[List[int]]
        :rtype: int
        """
        graph = {i:[] for i in range(1, N+1)}
        for t in trust:
            graph[t[0]].append(t[1])
            
        # check if the judge exists
        for k in graph:
            if len(graph[k]) == 0:
                # check if others trusts the judge
                judge = k                
                for person in graph:                    
                    if person != judge and judge not in graph[person]:
                        return -1
                return judge                    
                
        return -1