leetcode997 Find the Town Judge

"""
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
"""
class Solution:
    def findJudge(self, N, trust):
        degree = [0]*(N+1)    #(N+1)的原因是要存对应1~N个人的度数
        for i, j in trust:    #对每个trust对进行遍历
            degree[i] -= 1    #第i个人出度减1
            degree[j] += 1    #第j个人入度加1
        for i in range(1, N+1):  #对1到N+1个人进行遍历
            if degree[i] == N - 1: #找到出度为0，入度为N-1的人，即为town judge
                return i
        return -1

"""
开一个N+1长度的容器。
容器的索引是人的序号。
把对放入容器里。pair<出度，入度>。
遍历数组，更改每个人的出度和入度。
最后判断。出度为0，入度 为N-1的就是法官。
"""


#自练：
#针对图问题，将向量转变成二维矩阵的方法
import numpy as np
array = np.ones((4, 4))*0
N = 4
check = [[0 for i in range(N)] for j in range(N)]
trust = [[1, 3], [1, 4], [2, 3], [2, 4], [4, 3]]
for pair in trust:
    array[pair[0] - 1][pair[1] - 1] = 1
    check[pair[0] - 1][pair[1] - 1] = 1
print(array)
print(check)