LeetCode 210 Course Schedule II

最新推荐文章于 2025-08-08 11:49:43 发布

WANSNIM

最新推荐文章于 2025-08-08 11:49:43 发布

阅读量349

点赞数

CC 4.0 BY-SA版权

分类专栏： *LeetCode 文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/Codeblocksm/article/details/71170497

*LeetCode 专栏收录该内容

10 篇文章

订阅专栏

本文介绍了一种解决课程安排问题的算法实现。该问题要求根据课程总数及先修课程关系，找出一种有效的学习顺序。文章提供了使用拓扑排序方法的具体代码实现，并探讨了可能存在的多种正确顺序。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 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.

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 the correct course order is [0,1]

4, [[1,0],[2,0],[3,1],[3,2]]
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].

拓扑排序，难度Medium。

class Solution {
public:
    vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {

        int cnt[5100];  memset(cnt,0,sizeof(cnt));
        bool vis[5100];  memset(vis,false,sizeof(vis));
        vector<int> vet[5100];
        for(int i=0; i<prerequisites.size();i++)
        {
            pair<int,int> pr = prerequisites[i];
            vet[pr.second].push_back(pr.first);
            cnt[pr.first]++;
        }

        vector<int> ans;
        while(ans.size()<numCourses)
        {
            int u = -1;
            for(int i=0; i < numCourses; i++) if(cnt[i]==0 && !vis[i])  { u = i; break;} 
            if(u==-1 && ans.size() != numCourses) { ans.clear(); return ans;}
            ans.push_back(u); vis[u] = true;
            for(int i = 0; i<vet[u].size();i++) if(!vis[vet[u][i]]) cnt[vet[u][i]]--;
        }
        return ans;
    }
};