[leetcode] 207. Course Schedule 解题报告

最新推荐文章于 2024-01-21 17:16:08 发布

小榕流光

最新推荐文章于 2024-01-21 17:16:08 发布

阅读量1.8k

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode 拓扑排序有向图文章标签： leetcode 拓扑排序有向图

本文链接：https://blog.youkuaiyun.com/qq508618087/article/details/50748965

leetcode 同时被 3 个专栏收录

404 篇文章

订阅专栏

拓扑排序

4 篇文章

订阅专栏

有向图

2 篇文章

订阅专栏

该博客详细介绍了LeetCode中的207号问题——Course Schedule，讨论如何判断能否完成所有课程。通过解析题目、提供示例和讲解拓扑排序的概念，博主分享了使用DFS或BFS解决这个问题的方法，同时提供了具体的代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目链接：https://leetcode.com/problems/course-schedule/

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, is it possible for you to finish all courses?

For example:

2, [[1,0]]

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 course 1. So it is impossible.

Note:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.

click to show more hints.

Hints:

This problem is equivalent to finding if a cycle exists 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.

思路：拓扑排序的应用．拓扑顺序就像是在一个工程中每个小项目完成有向后顺序一样，而我们的任务就是找出这个完成的先后顺利的序列．

其方法就是每次将一个入度为０的结点删除，并且将其指向的结点入度减一．

代码如下：

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        if(numCourses == 0) return false;
        vector<int> indegree(numCourses, 0);
        unordered_map<int, multiset<int>> hash;
        for(auto val: prerequisites)
        {
            indegree[val.first]++;
            hash[val.second].insert(val.first);
        }
        for(int i = 0, j; i < numCourses; i++)
        {
            for(j = 0; j < numCourses; j++)
                if(indegree[j]==0) break;
            if(j==numCourses) return false;
            indegree[j]--;
            for(auto val: hash[j]) indegree[val]--;
        }
        return true;
    }
};