997. Find the Town Judge

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

class Solution {
    public int findJudge(int N, int[][] trust) {
        if(trust.length == 1) return trust[0][1];
        if(trust.length == 0) return 1;
        Set<Integer> set = new HashSet();
        for(int i = 0; i < trust.length; i++){
            set.add(trust[i][1]);
        }
        for(int i = 0; i < trust.length; i++){
            if(set.contains(trust[i][0])) set.remove(trust[i][0]);
        }
        int res = -1;
        int ind = 0;
        if(set.size() != 0) res =  set.iterator().next();
        for(int i = 0; i < trust.length; i++){
            if(trust[i][1] == res) ind++;
        }
        return ind == N-1 ? res : -1;
    }
}

普通方法得考虑很多边角问题比如数组空、1，不是所有其他人相信judge

用图就比较方便

https://leetcode.com/problems/find-the-town-judge/discuss/242938/JavaC%2B%2BPython-Directed-Graph

参考lee神

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

in-degree - out-degree = N - 1