# 997. Find the Town Judge 找到小镇的法官

@TOC

## # 题目描述

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] != trustiopen in new window
5. 1 <= trust[i][0], trustiopen in new window <= N

## # 题目大意

1. 法官谁也不信任
2. 每个人都信任这个人
3. 只有一个人满足条件1和2

## # 解题方法

### # 度

C++代码如下：

``````class Solution {
public:
int findJudge(int N, vector<vector<int>>& trust) {
vector<int> g(N + 1, 0); // in-degree - out-degree
for (auto t : trust) {
++g[t[1]];
--g[t[0]];
}
for (int i = 1; i <= N; ++i) {
if (g[i] == N - 1)
return i;
}
return -1;
}
};
``````

## # 日期

2019 年 2 月 24 日 —— 周末又结束了