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
Note:
1 <= N <= 1000
trust.length <= 10000
trust[i]
are all differenttrust[i][0] != trust[i][1]
1 <= trust[i][0], trust[i][1] <= N
题目理解:
给定一组人与人之间的信任关系,[a,b]表示a信任b。定义judge:judge这个人不相信任何人,但是被所有人相信。在给定的信任关系中,找出judge,如果不能找出,返回-1
解题思路:
记录所有人信任的数目和被人信任的数目,如果有一个人的信任数是0,而被信任数是N-1,那么他就是judge。如果没有这样的人,则找不出judge
class Solution {
public int findJudge(int N, int[][] trust) {
int[][] record = new int[N + 1][2];
for(int[] it : trust){
record[it[0]][0]++;
record[it[1]][1]++;
}
for(int i = 1; i < N + 1; i++)
if(record[i][0] == 0 && record[i][1] == N - 1)
return i;
return -1;
}
}