[Leetcode 207, Medium] Course Schedule

最新推荐文章于 2025-08-22 19:21:44 发布

simplify2012

最新推荐文章于 2025-08-22 19:21:44 发布

阅读量618

点赞数

CC 4.0 BY-SA版权

分类专栏： Leetcode_medium 文章标签： leetcode Graph Topological Sort

本文链接：https://blog.youkuaiyun.com/u012212811/article/details/45787425

Leetcode_medium 专栏收录该内容

99 篇文章

订阅专栏

探讨了如何判断在给定课程总数及先修课程约束条件下，是否能完成所有课程的问题。通过构建有向图并利用拓扑排序的方法来检测是否存在环路，从而确定能否成功完成所有课程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Problem:

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.

Analysis:

Solutions:

C++:

    bool canFinish(int numCourses, vector<pair<int, int> >& prerequisites) {
        if(numCourses == 0)
            return false;
            
        if(numCourses == 1)
            return true;
        
        int dep[numCourses] = {0};
        bool pre[numCourses][numCourses];
        for(int i = 0; i < numCourses; ++i) {
            for(int j = 0; j < numCourses; ++j) {
                pre[i][j] = false;
            }
        }
        
        int edge_num = 0;
        for(int i = 0; i < prerequisites.size(); ++i) {
            if(!pre[prerequisites[i].second][prerequisites[i].first]) {
                ++dep[prerequisites[i].first];
                pre[prerequisites[i].second][prerequisites[i].first] = true;
                ++edge_num;
            }
        }
        
        queue<int> no_pre_courses;
        for(int i = 0; i < numCourses; ++i) {
            if(dep[i] == 0)
                no_pre_courses.push(i);
        }
        
        if(no_pre_courses.empty())
            return false;

        while(!no_pre_courses.empty()) {
            int course = no_pre_courses.front();
            no_pre_courses.pop();
            
            for(int i = 0; i < numCourses; ++i) {
                if(pre[course][i]) {
                    if(--dep[i] == 0)
                        no_pre_courses.push(i);
                    pre[course][i] = false;
                    --edge_num;
                }
            }
        }
        
        return edge_num == 0;
    }

Java :

Python: