#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 <= 1000trust.length <= 10000trust[i]are all differenttrust[i][0] != trust[i][1]1 <= trust[i][0], trust[i][1] <= N
代码:
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;
}
};
#Leetcode# 997. Find the Town Judge的更多相关文章
- 【Leetcode_easy】997. Find the Town Judge
problem 997. Find the Town Judge solution: class Solution { public: int findJudge(int N, vector<v ...
- 【LeetCode】997. Find the Town Judge 解题报告(C++)
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 度 日期 题目地址:https://leetcode ...
- 【leetcode】997. Find the Town Judge
题目如下: In a town, there are N people labelled from 1 to N. There is a rumor that one of these people ...
- [Swift]LeetCode997. 找到小镇的法官 | 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 se ...
- LeetCode.997-找到镇法官(Find the Town Judge)
这是悦乐书的第373次更新,第400篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第234题(顺位题号是997).在一个城镇,有N个人从1到N标记.有传言说其中一个人是秘 ...
- LeetCode题解之 Find the Town Judge
1.题目描述 2.问题分析 使用map set数据结构. 3.代码 int findJudge(int N, vector<vector<int>>& trust) { ...
- LeetCode997. Find the Town Judge
题目 在一个小镇里,按从 1 到 N 标记了 N 个人.传言称,这些人中有一个是小镇上的秘密法官. 如果小镇的法官真的存在,那么: 小镇的法官不相信任何人. 每个人(除了小镇法官外)都信任小镇的法官. ...
- 算法与数据结构基础 - 图(Graph)
图基础 图(Graph)应用广泛,程序中可用邻接表和邻接矩阵表示图.依据不同维度,图可以分为有向图/无向图.有权图/无权图.连通图/非连通图.循环图/非循环图,有向图中的顶点具有入度/出度的概念. 面 ...
- Swift LeetCode 目录 | Catalog
请点击页面左上角 -> Fork me on Github 或直接访问本项目Github地址:LeetCode Solution by Swift 说明:题目中含有$符号则为付费题目. 如 ...
随机推荐
- 第36章 扩展授权 - Identity Server 4 中文文档(v1.0.0)
OAuth 2.0为令牌端点定义了标准授权类型,例如password,authorization_code和refresh_token.扩展授权是一种添加对非标准令牌颁发方案(如令牌转换,委派或自定义 ...
- Entity Framework 框架
微软官方提供的ORM技术的实现就是EF(Entity Framework)框架.EF的模式有三种分别是:Database First 数据库先行 ,Model First 模型先行 , Code F ...
- 浅谈_依赖注入 asp.net core
1.1什么是依赖 我们先看下图 可以简单理解,一个HomeController类使用到了DBContext类,而这种关系是有偶然性,临时性,弱关系的,但是DBContext的变化会影响到HomeCon ...
- 814-Binary Tree Pruning
Description: We are given the head node root of a binary tree, where additionally every node’s value ...
- Spring笔记03_AOP
目录 1. AOP 1.1 AOP介绍 1.1.1 什么是AOP 1.1.2 AOP实现原理 1.1.3 AOP术语[掌握] 1.2 AOP的底层实现(了解) 1.2.1 JDK动态代理 1.2.2 ...
- 前端js 实现文件下载
https://www.zhangxinxu.com/wordpress/2017/07/js-text-string-download-as-html-json-file/ 侵删 1.H5 down ...
- #WEB安全基础 : HTML/CSS | 0x5a标签拓展和class、id属性的使用
a标签不只是能链接到其他网页,也能链接到网页中的元素 class属性让你用CSS对特定的元素进行修饰 这些是一个网页设计者的有力武器 这节课还是一个index.html文件 以下是代码 <h ...
- 在AndroidStudio上使用AddressSanitizer
在AndroidStudio上使用AddressSanitizer AddressSanitizer是Google主导的一个开源内存问题检测工具.现在也开始支持Android平台,且受Google推荐 ...
- 从零学习Fluter(四):Flutter中ListView组件系列详展
今天继续研究了一些Flutter,主要时关于ListVIew那一块的东西,有 SingleChildScrollViewListViewGridViewCustomScrollView 感觉Flutt ...
- 第三篇 Html-label标签
label标签 用户获取文字,使得关联的标签获取光标 <!DOCTYPE html> <html lang="en"> <head> <m ...