LeetCode:Find the Town Judge（Python3实现）

题目描述

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:
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

输入：N = 2, trust = [[1,2]]
输出：2

输入：N = 3, trust = [[1,3],[2,3]]
输出：3

输入：N = 3, trust = [[1,3],[2,3],[3,1]]
输出：-1

输入：N = 3, trust = [[1,2],[2,3]]
输出：-1

输入：N = 4, trust = [[1,3],[1,4],[2,3],[2,4],[4,3]]
输出：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

该题目考察的是离散数学中的有向图，[a,b]指的是一条从a指向b的有向边，换言之，题目意思就是找到一个所有点都指向它的点，即它的入度为N-1,出度为0，而且在有向图中这个点是唯一的。

方法一

建立一个degree数组存储1到N各节点的入度和出度，循环遍历trust找到每个数的出度和入度，直到找到入度为N-1,出度为0的点，否则就返回-1。这种方法复杂度为**O(Nlen(trust))***,运行较慢。

代码

class Solution:
    def findJudge(self, N: int, trust: List[List[int]]) -> int:
        degree=[]
        for i in range(1,N+1):
            indegree=0
            outdegree=0
            for j in range(len(trust)):
                if trust[j][0]==i:
                    outdegree+=1;
                if trust[j][1]==i:
                    indegree+=1;
            if indegree==N-1 and outdegree==0:
                return i
        return -1

运行结果

方法二

七桥问题解法直接计算各个节点的入度和出度的差值，得到的差值为N-1的节点为法官，这个方法复杂度为***O(N+len(trust))***，速度明显提升。
参考博客：https://blog.youkuaiyun.com/qq_17550379/article/details/87903873

代码

class Solution:
    def findJudge(self, N: int, trust: List[List[int]]) -> int:
        count = [0] * (N + 1)
        for i, j in trust:
            count[i] -= 1
            count[j] += 1
        for i in range(1, N + 1):
            if count[i] == N - 1:
                return i
        return -1

运行结果

总结

此博客为博主第一篇博客，希望对在刷LeetCode的各位程序猿们有所帮助。