leetcode 997. Find the Town Judge（python）

描述

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
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

解析

根据题意，要找出隐藏在城镇中的法官，法官必须满足其他人都相信他，他不相信任何人。现在给出了 trust 列表表示 a 相信 b ，这样就可以对每个人进行计数。用长度为 N+1 的列表 d 表示所有人（索引为 0 的元素无用，其他索引代表城镇的每个人），然后遍历 trust 中的每个子列表 [a,b] ，d[a]-=1 表示他如果去相信其他人表示他肯定不是法官， d[b]+=1 表示他被人相信则可能是法官，遍历结束之后，只要 d 中的位置的计数为 N-1 ，就表示其为法官，将其索引返回，否则返回 -1 。

解答

 class Solution(object):
    def findJudge(self, N, trust):
        """
        :type N: int
        :type trust: List[List[int]]
        :rtype: int
        """
        d = [0] * (N+1)
        for a,b in trust:
            d[a] -= 1
            d[b] += 1
        for person in range(1, N + 1):
            if d[person] == N - 1:
                return person
        return -1           	      

运行结果

Runtime: 596 ms, faster than 96.12% of Python online submissions for Find the Town Judge.
Memory Usage: 18.8 MB, less than 22.99% of Python online submissions for Find the Town Judge.

原题链接：https://leetcode.com/problems/find-the-town-judge/

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