例如,想要学习课程0,你需要先完成课程1,我们用一个匹配来表示:[0,1]。 返回你为了学完所有课程所安排的学习顺序。可能会有多个正确的顺序,你只要返回任意一种就可以了。如果不可能完成所有课程,返回一个空数组。 示例1: 输入:numCourses = 2, prerequisites = [[1,0]]输出:[0,1]解释:总共有 2 门课程。
boolLeetCode::canFinish(intnumCourses, vector<pair<int,int>>&prerequisites){if(numCourses <1)returntrue;//没有课程if(!prerequisites.size())returntrue;//没有依赖关系vector<vector<int>>courses(numCourses);//第i个课程依赖于courses[i]里的课程for(size_t i =0; i < prerequisites.size(); i+...
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 - Course Schedule II There are a total of n courses you have to take labelled from 0 to n - 1. Some courses may have prerequisites, for example, if prerequisites[i] = [ai, bi] this means you must take the course bi before the course ai. Given the total number of courses...
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...
建议和leetcode 210. Course Schedule II 拓扑排序 + HashSet和leetcode 630. Course Schedule III 课程调度 + 贪心算法 一起学习 需要注意的是,由于这道题更新了测试用例所以原先的代码可能出现问题,情况是这样的,记录后继结点要使用vector,不要使用set,因为假如存在(1,9)和(1,9)多次出现,这样就会出现问题,所...
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 ...
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) { ...
[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: [0,1] Given the total number ...
add(n); } } int[] res = new int[numCourses]; int j = 0; while(!zeroQueue.isEmpty()) { Node cur = zeroQueue.poll(); res[j++] = cur.value; for(Node next: cur.next) { next.in--; if(next.in == 0) { zeroQueue.add(next); } } } return j == numCourses ? res: ...