[LeetCode] 210. Course Schedule II 课程清单之二
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 of courses and a list of prerequisite pairs, return the ordering of courses you should take to finish all courses.
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.
Example 1:
Input: 2, [[1,0]]
Output:[0,1]
Explanation: There are a total of 2 courses to take. To take course 1 you should have finished
course 0. So the correct course order is[0,1] .
Example 2:
Input: 4, [[1,0],[2,0],[3,1],[3,2]]
Output:[0,1,2,3] or [0,2,1,3]
Explanation: There are a total of 4 courses to take. To take course 3 you should have finished both
courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0.
So one correct course order is[0,1,2,3]
. Another correct ordering is[0,2,1,3] .
Note:
- The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
- You may assume that there are no duplicate edges in the input prerequisites.
- This problem is equivalent to finding the topological order in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
- Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
- Topological sort could also be done via BFS.
这题是之前那道 Course Schedule 的扩展,那道题只让我们判断是否能完成所有课程,即检测有向图中是否有环,而这道题我们得找出要上的课程的顺序,即有向图的拓扑排序 Topological Sort,这样一来,难度就增加了,但是由于我们有之前那道的基础,而此题正是基于之前解法的基础上稍加修改,我们从 queue 中每取出一个数组就将其存在结果中,最终若有向图中有环,则结果中元素的个数不等于总课程数,那我们将结果清空即可。代码如下:
class Solution {
public:
vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {
vector<int> res;
vector<vector<int> > graph(numCourses, vector<int>());
vector<int> in(numCourses, );
for (auto &a : prerequisites) {
graph[a.second].push_back(a.first);
++in[a.first];
}
queue<int> q;
for (int i = ; i < numCourses; ++i) {
if (in[i] == ) q.push(i);
}
while (!q.empty()) {
int t = q.front();
res.push_back(t);
q.pop();
for (auto &a : graph[t]) {
--in[a];
if (in[a] == ) q.push(a);
}
}
if (res.size() != numCourses) res.clear();
return res;
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/210
类似题目:
参考资料:
https://leetcode.com/problems/course-schedule-ii/
LeetCode All in One 题目讲解汇总(持续更新中...)
[LeetCode] 210. Course Schedule II 课程清单之二的更多相关文章
- [LeetCode] Course Schedule II 课程清单之二
There are a total of n courses you have to take, labeled from 0 to n - 1. Some courses may have prer ...
- [LeetCode] 210. Course Schedule II 课程安排II
There are a total of n courses you have to take, labeled from 0 to n - 1. Some courses may have prer ...
- Java for LeetCode 210 Course Schedule II
There are a total of n courses you have to take, labeled from 0 to n - 1. Some courses may have prer ...
- LeetCode 210. Course Schedule II(拓扑排序-求有向图中是否存在环)
和LeetCode 207. Course Schedule(拓扑排序-求有向图中是否存在环)类似. 注意到.在for (auto p: prerequistites)中特判了输入中可能出现的平行边或 ...
- Leetcode 210 Course Schedule II
here are a total of n courses you have to take, labeled from 0 to n - 1. Some courses may have prere ...
- [leetcode]210. Course Schedule II课程表II
There are a total of n courses you have to take, labeled from 0 to n-1. Some courses may have prereq ...
- (medium)LeetCode 210.Course Schedule II
There are a total of n courses you have to take, labeled from 0 to n - 1. Some courses may have prer ...
- [LeetCode] Course Schedule III 课程清单之三
There are n different online courses numbered from 1 to n. Each course has some duration(course leng ...
- 【LeetCode】210. Course Schedule II
Course Schedule II There are a total of n courses you have to take, labeled from 0 to n - 1. Some co ...
随机推荐
- python yield: send, close, throw
send 1. yield可以产出值,可以接收值 2. 在调用send发送非none值之前,我们必须启动一次生成器, 方式有两种 a. gen.send(None) b. next(gen) def ...
- Entity Framework Core 练习参考
项目地址:https://gitee.com/dhclly/IceDog.EFCore 项目介绍 对 Microsoft EntityFramework Core 框架的练习测试 参考文档教程 官方文 ...
- 在 EF Core 中 Book 实体在新增、修改、删除时,给 LastUpdated 字段赋值。
直接贴代码: public class MenusContext : DbContext { public static class ColumnNames { public const string ...
- Linux iSCSI 磁盘共享管理
Linux iSCSI 磁盘共享管理 iSCSI 服务是通过服务端(target)与客户端(initiator)的形式来提供服务.iSCSI 服务端用于存放存储源的服务器,将磁盘空间共享给客户使用,客 ...
- JVM的监控工具之jconsole
JConsole(Java Monitoring and Management Console)是一种基于JMX的可视化监视.管理工具.管理的是什么?管理的是监控信息.永久代的使用信息.类加载等等 如 ...
- CentOS安装etcd和flannel实现Docker跨物理机通信
1.安装etcd yum install etcd systemctl stop etcd systemctl start etcd systemctl status etcd systemctl e ...
- vue入门案例
1.技术在迭代,有时候你为了生活没有办法,必须掌握一些新的技术,可能你不会或者没有时间造轮子,那么就先把利用轮子吧. <!DOCTYPE html> <html> <he ...
- .Net常见的IOC框架及AOP框架
IOC框架 Unity:微软patterns&practicest团队开发的IOC依赖注入框架,支持AOP横切关注点. MEF(Managed Extensibility Framework) ...
- scrapy学习笔记(二)框架结构工作原理
scrapy结构图: scrapy组件: ENGINE:引擎,框架的核心,其它所有组件在其控制下协同工作. SCHEDULER:调度器,负责对SPIDER提交的下载请求进行调度. DOWNLOADER ...
- CentOS 6.9安装MySQL 5.6 (使用yum安装)
CentOS 6.9安装MySQL 5.6 (使用yum安装) 移除CentOS默认的mysql-libs [root@test01 srv]# whereis mysqlmysql: /usr/li ...