存在a->b,又存在b->c,那么增加边a->c 如果新增边c->a,发现已有a->c,在矩阵中表现为对称位置为true,则存在回路。 注:数组比vector速度快。 classSolution {public:boolcanFinish(intnumCourses, vector<pair<int,int>>&prerequisites) {bool**preG; preG=newbool*[numCourses];for(inti =0; i < numC...
LeetCode - Course Schedule 解题报告 以前从来没有写过解题报告,只是看到大肥羊河delta写过不少。最近想把写博客的节奏给带起来,所以就挑一个比较容易的题目练练手。 原题链接 https://leetcode.com/problems/course-schedule/ 题目大意 有n个课程,编号分别是0到n-1。我们的目标是修完所有课程。然而有些课程...
1 classSolution{ 2 public: 3 vector<int>findOrder(intnumCourses,vector<vector<int>>&prerequisites) { 4 5 } 6 }; 您必须登录后才能提交解答! © 2025 领扣网络(上海)有限公司版权所有 沪公网安备 31011502007040号 沪ICP备17051546号
【LeetCode】207. Course Schedule 解题报告(Python) 作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 题目地址:https://leetcode.com/problems/course-schedule/description/ 题目描述: There are a total of n courses you have to take, labeled from0ton-1. Some courses may have pr...
There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array. 使用拓扑排序即可,代码如下: class Solution { public: vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) { ...
题目地址:https://leetcode-cn.com/problems/course-schedule-iv/ 题目描述 你总共需要上n门课,课程编号依次为0到n-1。 有的课会有直接的先修课程,比如如果想上课程0,你必须先上课程1,那么会以[1,0]数对的形式给出先修课程数对。 给你课程总数n和一个直接先修课程数对列表prerequisite和一个查询对列表queri...
Can you solve this real interview question? Course Schedule IV - There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must ta
Leetcode 207. Course Schedule (Python) 仗剑走天涯 语音信号处理识别刚入行 来自专栏 · 思考的艺术 1 人赞同了该文章 There are a total of n courses you have to take, labeled from 0 to n-1. Some courses may have prerequisites, for example to take course 0 you have to first take course ...
[LeetCode]CourseSchedule书影博客[LeetCode]CourseSchedule 书影博客 题目描述: There are a total of n courses you have to take, labeled from 0 to n - 1. Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0...
最后判断一下是否所有节点都经过了该流程,若处理节点个数没有达到numCourses返回空数组即可。 复杂度分析 时间复杂度: O(N): 取决于处理的节点个数。 空间复杂度: O(N): 取决于处理的节点个数。 代码 class Solution { class Node { int in; int value; List<Node> next; public Node(int value) { in...