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 <= 1000trust.length <= 10000trust[i]are all differenttrust[i][0] != trust[i][1]1 <= trust[i][0], trust[i][1] <= N
aclass Solution {
public int findJudge(int N, int[][] trust) {
Map<Integer, Integer> hmap = new HashMap<>();
if(trust.length == 0){
return 1;
}
for(int i=0; i<trust.length; i++){
Integer people = trust[i][0];
Integer tpeople = trust[i][1];
hmap.put(people, hmap.getOrDefault(people,0)-1);
hmap.put(tpeople, hmap.getOrDefault(tpeople,0)+1);
}
for(Map.Entry<Integer, Integer> item : hmap.entrySet()){
Integer tpeople = item.getKey();
Integer tcount = item.getValue();
if(tcount == N - 1){
return tpeople;
}
}
return -1;
}
}
class Solution {
public int findJudge(int N, int[][] trust) {
int[] trusted = new int[N];
for(int i=0; i<trust.length; i++){
int people = trust[i][0];
int tpeople = trust[i][1];
trusted[tpeople-1]++;
trusted[people-1]--;
}
for(int i=0; i<trusted.length; i++){
if(trusted[i] == N - 1){
return i+1;
}
}
return -1;
}
}[문제] https://leetcode.com/problems/find-the-town-judge/
