题目地址 https://leetcode.com/problems/course-schedule-ii/ 题目大意 https://leetcode-cn.com/problems/course-schedule-ii 解题思路 有向图的拓扑排序,采用BFS解决。对每组数字,如[1,0], 实际可表示为一条从结点0指向结点1的边,对记录下每个结点的入度数。循环遍历每个结点,如果入度为0,表示没有需要上的...
Well, this problem is spiritually similar to toCourse Schedule. You only need to store the nodes in the order you visit into a vector during BFS or DFS. Well, for DFS, a final reversal is required. BFS 1classSolution {2public:3vector<int> findOrder(intnumCourses, vector<pair<int,int>...
贡献者:LeetCode 相关标签 深度优先搜索广度优先搜索图拓扑排序 相似题目 课程表火星词典最小高度树序列重建课程表 III并行课程从给定原材料中找到所有可以做出的菜给定条件下构造矩阵通过移动项目到空白区域来排序数组 1 classSolution{ 2 public: 3 vector<int>findOrder(intnumCourses,vector<vector<int>>&prerequisit...
Leetcode】210.course-schedule-ii题目地址 https://leetcode.com/problems/course-schedule-ii/ 题目大意 https://leetcode-cn.com/problems/course-schedule-ii 解题思路 有向图的拓扑排序,采用BFS解决。对每组数字,如[1,0], 实际可表示为一条从结点0指向结点1的边,对记录下每个结点的入度数。循环遍历每个...
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】Course Schedule II 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...
[leetcode] 210. Course Schedule II Description 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]...
参考上题,Java for LeetCode 207 Course Schedule【Medium】 修改下代码即可,JAVA实现如下: publicint[]findOrder(intnumCourses,int[][]prerequisites){ int[]res=newint[numCourses]; List<Set<Integer>>posts=newArrayList<Set<Integer>>(); for(inti=0;i<numCourses;i++) ...
我在leetcode 210上成功AC的代码,为何在linCode 616上卡在了10001的case,报错temsig=11,是因为占用空间太多? class Solution: # @param {int} numCourses a total of n courses # @param {int[][]} prerequisites a list of prerequisite pairs # @return {int[]} the course order neighbors = None vis...
class Solution { class Node { int in; int value; List<Node> next; public Node(int value) { in = 0; this.value = value; next = new ArrayList<>(); } } public int[] findOrder(int numCourses, int[][] prerequisites) { Map<Integer, Node> nodeMap = new HashMap<>(); for(int ...