leetcode

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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

小镇里有 n 个人，按从 1 到 n 的顺序编号。传言称，这些人中有一个暗地里是小镇法官。

如果小镇法官真的存在，那么：

• 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 。

给你一个数组 trust ，其中 trust[i] = [ai, bi] 表示编号为 ai 的人信任编号为 bi 的人。

如果小镇法官存在并且可以确定他的身份，请返回该法官的编号；否则，返回 -1 。

解题思路

入度和出度统计

• 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
}