LeetCode 997. Find the Town Judge

Posted 2021-03-22 Dylan_Java_NYC

原题链接在这里：https://leetcode.com/problems/find-the-town-judge/

题目：

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:

1. The town judge trusts nobody.
2. Everybody (except for the town judge) trusts the town judge.
3. 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. 1 <= N <= 1000
2. trust.length <= 10000
3. trust[i] are all different
4. trust[i][0] != trust[i][1]
5. 1 <= trust[i][0], trust[i][1] <= N

题解：

Could find a candidate first with its trust set is empty.

Then iterate trust sets again to see if any other people\'s trust sets contains candidate, if not return -1.

Time Compleixty: O(N).

Space: O(N).

AC Java:

 1 class Solution {
 2     public int findJudge(int N, int[][] trust) {
 3         if(N < 2){
 4             return N;
 5         }
 6         
 7         HashSet<Integer> [] trustSets = new HashSet[N+1];
 8         for(int i = 1; i<=N; i++){
 9             trustSets[i] = new HashSet<>();
10         }
11         
12         for(int [] t : trust){
13             trustSets[t[0]].add(t[1]);
14         }
15         
16         int c = -1;
17         for(int i = 1; i<=N; i++){
18             if(trustSets[i].isEmpty()){
19                 c = i;
20             }
21         }
22         
23         for(int i = 1; i<=N; i++){
24             if(i != c && !trustSets[i].contains(c)){
25                 return -1;
26             }
27         }
28         
29         return c;
30     }
31 }

We could count the trusts.

For each trust array t, t[0]--, t[1]++. And check if there is one count == N-1.

Time Complexity: O(N).

Space: O(N).

AC Java:

 1 class Solution {
 2     public int findJudge(int N, int[][] trust) {
 3         if(N < 2){
 4             return N;
 5         }
 6         
 7         int [] count = new int[N+1];
 8         for(int [] t : trust){
 9             count[t[0]]--;
10             count[t[1]]++;
11         }
12         
13         for(int i = 1; i<=N; i++){
14             if(count[i] == N - 1){
15                 return i;
16             }
17         }
18         
19         return -1;
20     }
21 }

类似Find the Celebrity.