leetcode997 Find the Town Judge

 1 """
 2 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.
 3 If the town judge exists, then:
 4     The town judge trusts nobody.
 5     Everybody (except for the town judge) trusts the town judge.
 6     There is exactly one person that satisfies properties 1 and 2.
 7 You are given trust, an array of pairs trust[i] = [a, b] representing that the person labelled a trusts the person labelled b.
 8 If the town judge exists and can be identified, return the label of the town judge.  Otherwise, return -1.
 9 Example 1:
10 Input: N = 2, trust = [[1,2]]
11 Output: 2
12 Example 2:
13 Input: N = 3, trust = [[1,3],[2,3]]
14 Output: 3
15 Example 3:
16 Input: N = 3, trust = [[1,3],[2,3],[3,1]]
17 Output: -1
18 Example 4:
19 Input: N = 3, trust = [[1,2],[2,3]]
20 Output: -1
21 Example 5:
22 Input: N = 4, trust = [[1,3],[1,4],[2,3],[2,4],[4,3]]
23 Output: 3
24 """
25 class Solution:
26     def findJudge(self, N, trust):
27         degree = [0]*(N+1)    #(N+1)的原因是要存对应1~N个人的度数
28         for i, j in trust:    #对每个trust对进行遍历
29             degree[i] -= 1    #第i个人出度减1
30             degree[j] += 1    #第j个人入度加1
31         for i in range(1, N+1):  #对1到N+1个人进行遍历
32             if degree[i] == N - 1: #找到出度为0，入度为N-1的人，即为town judge
33                 return i
34         return -1
35 
36 """
37 开一个N+1长度的容器。
38 容器的索引是人的序号。
39 把对放入容器里。pair<出度，入度>。
40 遍历数组，更改每个人的出度和入度。
41 最后判断。出度为0，入度 为N-1的就是法官。
42 """
43 
44 
45 #自练：
46 #针对图问题，将向量转变成二维矩阵的方法
47 import numpy as np
48 array = np.ones((4, 4))*0
49 N = 4
50 check = [[0 for i in range(N)] for j in range(N)]
51 trust = [[1, 3], [1, 4], [2, 3], [2, 4], [4, 3]]
52 for pair in trust:
53     array[pair[0] - 1][pair[1] - 1] = 1
54     check[pair[0] - 1][pair[1] - 1] = 1
55 print(array)
56 print(check)