0997. Find the Town Judge

# 997. Find the Town Judge#

## 题目 #

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.

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 <= 10000
• trust[i].length == 2
• All the pairs of trust are unique.
• ai != bi
• 1 <= ai, bi <= n

## 题目大意 #

• 小镇法官不会信任任何人。
• 每个人（除了小镇法官）都信任这位小镇法官。
• 只有一个人同时满足属性 1 和属性 2 。

## 解题思路 #

• 被人信任定义为入度, 信任别人定义为出度
• 如果 1-n 之间有数字 x 的入度为 n - 1，出度为 0，则返回 x

## 代码 #

``````package leetcode

func findJudge(n int, trust [][]int) int {
if n == 1 && len(trust) == 0 {
return 1
}
judges := make(map[int]int)
for _, v := range trust {
judges[v[1]] += 1
}
for _, v := range trust {
if _, ok := judges[v[0]]; ok {
delete(judges, v[0])
}
}
for k, v := range judges {
if v == n-1 {
return k
}
}
return -1
}
``````

Apr 8, 2023