LeetCode 997. Find the Town Judge

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:

class Solution {
    public int findJudge(int N, int[][] trust) {
        if(N < 2){
            return N;
        }
        
        HashSet<Integer> [] trustSets = new HashSet[N+1];
        for(int i = 1; i<=N; i++){
            trustSets[i] = new HashSet<>();
        }
        
        for(int [] t : trust){
            trustSets[t[0]].add(t[1]);
        }
        
        int c = -1;
        for(int i = 1; i<=N; i++){
            if(trustSets[i].isEmpty()){
                c = i;
            }
        }
        
        for(int i = 1; i<=N; i++){
            if(i != c && !trustSets[i].contains(c)){
                return -1;
            }
        }
        
        return c;
    }
}

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:

class Solution {
    public int findJudge(int N, int[][] trust) {
        if(N < 2){
            return N;
        }
        
        int [] count = new int[N+1];
        for(int [] t : trust){
            count[t[0]]--;
            count[t[1]]++;
        }
        
        for(int i = 1; i<=N; i++){
            if(count[i] == N - 1){
                return i;
            }
        }
        
        return -1;
    }
}

AC C++:

class Solution {
public:
    int findJudge(int n, vector<vector<int>>& trust) {
        vector<int> map(n+1, 0);
        for(auto& t : trust){
            map[t[0]]--;
            map[t[1]]++;
        }

        for(int i = 1; i <= n; i++){
            if(map[i] == n - 1){
                return i;
            }
        }

        return -1;
    }
};

类似Find the Celebrity.