997.Find the Town Judge
===

###### tags: `Easy`,`Array`,`Hash Table`,`Graph`

[997. Find the Town Judge](https://leetcode.com/problems/find-the-town-judge/)

### 題目描述

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]` = [$a_i$, $b_i$] representing that the person labeled $a_i$ trusts the person labeled $b_i$.

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

### 範例

**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` <= 10^4^
* `trust[i].length` == 2
* All the pairs of `trust` are **unique**.
* $a_i$ != $b_i$
* 1 <= $a_i$, $b_i$ <= `n`

### 解答
#### C++

class Solution {
public:
    int findJudge(int n, vector<vector<int>>& trust) {
        vector<int> a(n), b(n);
        for (auto& t : trust) {
            a[t[0] - 1]++;
            b[t[1] - 1]++;
        }
        for (int i = 0; i < n; i++) {
            if (a[i] == 0 && b[i] == n - 1) {
                return i + 1;
            }
        }
        return -1;
    }
};

> [name=Yen-Chi Chen][time=Tue, Jan 24, 2023]

### Reference

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