LeetCode 997. 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:

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 an array trust where trust[i] = [ai, bi] representing that the person labeled ai trusts the person labeled bi.

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

---

就是给了一个长度为n的数组的数组[a, b]，表示a信任b，要找出唯一一个人，他不信任任何人，且其他所有人都信任他。其实看完题就想到了graph，但已经把graph全忘光了，于是就自己哼哧哼哧写了个brute force解法。

my brute force solution:

先把所有人放到一个set里，把出现在第一个的人remove掉，最后如果剩且只剩下一个人，这个人就是possible judge，否则return -1。然后再看一遍其他所有人是不是都trust这个possible judge。

class Solution {
    public int findJudge(int n, int[][] trust) {
        Set set = new HashSet<>();
        for (int i = 1; i <= n; i++) {
            set.add(i);
        }
        for (int[] i : trust) {
            set.remove(i[0]);
        }
        if (set.size() != 1) {
            return -1;
        }
        Iterator it = set.iterator();
        int judge = it.next();
        int[] all = new int[n + 1];
        all[0] = 1;
        all[judge] = 1;
        for (int[] i : trust) {
            if (i[1] == judge) {
                all[i[0]] = 1;
            }
        }
        for (int i : all) {
            if (i == 0) {
                return -1;
            }
        }
        return judge;
    }
}

然后看了解答，虽然是用的graph的思想但其实挺简单的，相当于judge的in degree是n -1，out degree是0，所以in - out = n -1。怎么保证没有其他的情况这个我还没想明白。如果分开计算in degree和out degree的话需要两个数组，但计算差值只需要一个数组。in degree（作为b）时++，out degree（作为a）时--。

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