输入:numCourses = 2, prerequisites = [[1,0]]输出:[0,1]解释:总共有 2 门课程。要学习课程 1,你需要先完成课程 0。因此,正确的课程顺序为[0,1] 。 示例2: 输入:numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]]输出:[0,2,1,3]解释:总共有 4 门课程。要学习课程 3,...
There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible. 2, [[1,0],[0,1]] There are a total of 2 courses to take. To take course 1 you should have finished course 0, and to take course 0 you should also have finished...
This is my reference:https://leetcode.com/discuss/35605/two-ac-solution-in-java-using-bfs-and-dfs-with-explanation BFS(please note the comment): classSolution {public: vector<int> findOrder(intnumCourses, vector<pair<int,int>>&prerequisites) { vector<int>ret; map<int,int>counts;for(inti...
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) { vector<int> res; ...
https://leetcode.com/problems/course-schedule-ii/ 题目: 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] ...
class Solution { public: vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) { vector<int> res; vector<vector<int>> graph(numCourses,vector<int>(0)); vector<int> in(numCourses,0); for(auto a: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 ...
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
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...
210. Course Schedule II vector<int>findOrder(intnumCourses,vector<vector<int>>&prerequisites){vector<int>ret;vector<vector<int>>dependency(numCourses,vector<int>());for(constauto&p:prerequisites){intleft=p[0];intright=p[1];dependency[left].push_back(right);}deque<int>stack;for(inti=0;i...