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

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

可以看作信任的關係是一個有向圖，建立一個計算in/out degree的counter array，看看每個人的indegree - outdegree是否為N - 1。

// Directed graph with counting degrees
int findJudge(int N, vector<vector<int>>& trust) { // time: O(T + N); space: O(N)
    vector<int> degree(N + 1, 0);
    for (vector<int>& t : trust) {
        --degree[t[0]], ++degree[t[1]];
    }
    for (int i = 1; i <= N; ++i) {
        if (degree[i] == N - 1) return i;
    }
    return -1;
}

277. Find the Celebrity