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  • 【leetcode】997. Find the Town Judge

    题目如下:

    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

    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

    解题思路:创建一个长度等于N的数组stat,记stat[i]为第i+1个人被别人信任的次数与信任别人的次数的差值,如果恰好等于N-1,则表明这个人就是法官。

    代码如下:

    class Solution(object):
        def findJudge(self, N, trust):
            """
            :type N: int
            :type trust: List[List[int]]
            :rtype: int
            """
            stat = [0] * N
            for x,y in trust:
                stat[x-1] -= 1
                stat[y-1] += 1
            for i,v in enumerate(stat):
                if v == N-1:
                    return i+1
            return -1
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  • 原文地址:https://www.cnblogs.com/seyjs/p/10426072.html
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