本题用两个数组分别记录每个结点的入度和出度。如果结点入度位N-1且出度为0，则该结点是法官。
JAVA
class Solution { public int findJudge(int N, int[][] trust) { if(trust.length < N-1) return -1; int[] indegrees = new int[N+1]; int[] outdegrees = new int[N+1]; for(int[] relation : trust){ outdegrees[relation[0]]++; indegrees[relation[1]]++; } for(int i = 1; i <= N; i++){ if(outdegrees[i] == 0 && indegrees[i] == N-1) return i; } return -1; } }
Python3
class Solution: def findJudge(self, N: int, trust: List[List[int]]) -> int: indegrees = [0]*(N+1) outdegrees = [0]*(N+1) for i, j in trust: outdegrees[i] += 1 indegrees[j] += 1 for i in range(1, N+1): #对1到N+1个人进行遍历 if indegrees[i] == N - 1 and outdegrees[i] == 0: #找到出度为0，入度为N-1的人，即为town judge return i return -1
May LeetCoding Challenge10 之 入度出度
原文：https://www.cnblogs.com/yawenw/p/12864234.html