leetcode-997-easy

Find the Town Judge

In a town, there are n people labeled 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 an array trust where trust[i] = [ai, bi] representing that the person labeled ai trusts the person labeled bi. If a trust relationship does not exist in trust array, then such a trust relationship does not exist.

Return the label of the town judge if the town judge exists and can be identified, or return -1 otherwise.

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
Constraints:

1 <= n <= 1000
0 <= trust.length <= 104
trust[i].length == 2
All the pairs of trust are unique.
ai != bi
1 <= ai, bi <= n

思路一：用 map 和 set 存储值，最后遍历判断

    public int findJudge(int n, int[][] trust) {
        if (trust.length == 0) return n == 1 ? 1 : -1;

        Map<Integer, Integer> map = new HashMap<>();
        Set<Integer> set = new HashSet<>();
        for (int[] t : trust) {
            map.compute(t[1], (k, v) -> v == null ? 1 : v + 1);
            set.add(t[0]);
        }

        for (Map.Entry<Integer, Integer> e : map.entrySet()) {
            if (e.getValue() == n - 1 && !set.contains(e.getKey())) {
                return e.getKey();
            }
        }

        return -1;
    }