#Leetcode# 997. Find the Town Judge
https://leetcode.com/problems/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:
- 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 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 <= N <= 1000
trust.length <= 10000
trust[i]
are all differenttrust[i][0] != trust[i][1]
1 <= trust[i][0], trust[i][1] <= N
代码:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 | class Solution { public : vector< int > v[1010]; int vis[1010]; int findJudge( int N, vector<vector< int >>& trust) { int n = trust.size(); for ( int i = 0; i < n; i ++) { int a = trust[i][0], b = trust[i][1]; v[b].push_back(a); vis[a] = 1; } int cnt = 0, temp = 0; for ( int i = 1; i <= N; i ++) { if (vis[i] == 0 && v[i].size() == N - 1) { cnt ++; temp = i; } } if (cnt == 1) return temp; return -1; } }; |
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】凌霞软件回馈社区,博客园 & 1Panel & Halo 联合会员上线
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】博客园社区专享云产品让利特惠,阿里云新客6.5折上折
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步
· 为什么说在企业级应用开发中,后端往往是效率杀手?
· 用 C# 插值字符串处理器写一个 sscanf
· Java 中堆内存和栈内存上的数据分布和特点
· 开发中对象命名的一点思考
· .NET Core内存结构体系(Windows环境)底层原理浅谈
· 为什么说在企业级应用开发中,后端往往是效率杀手?
· DeepSeek 解答了困扰我五年的技术问题。时代确实变了!
· 本地部署DeepSeek后,没有好看的交互界面怎么行!
· 趁着过年的时候手搓了一个低代码框架
· 推荐一个DeepSeek 大模型的免费 API 项目!兼容OpenAI接口!