[leetcode] Course Schedule II

最新推荐文章于 2024-09-11 04:40:53 发布

原创最新推荐文章于 2024-09-11 04:40:53 发布 · 349 阅读

0 ·

CC 4.0 BY-SA版权

Leetcode 专栏收录该内容

222 篇文章

订阅专栏

本文深入探讨了如何解决课程调度问题，具体通过LeetCode平台的一个实例来展示解决步骤和思路，包括如何利用图论知识进行课程依赖分析，并最终找到一个有效的课程完成顺序。

From : https://leetcode.com/problems/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].

class Solution { 
public: 
    vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) { 
        vector<int> ans; 
        vector<set<int>> posts(numCourses, set<int>()); 
        vector<int> preNums(numCourses, 0); 
        for(auto &ab : prerequisites) posts[ab.second].insert(ab.first); 
        for(int i=posts.size()-1; i>=0; i--) { 
            for(auto &nd : posts[i]) { 
                preNums[nd]++; 
            } 
        } 
        int num = numCourses; 
        queue<int> q; 
        for(int i=0; i<num; i++) { 
            if(preNums[i] == 0) q.push(i); 
        }
        while(!q.empty()) { 
            int k = q.front(); 
            q.pop();
			ans.push_back(k);
            for(auto &a : posts[k]) { 
                if(--preNums[a] == 0) { 
                    q.push(a);
                } 
            } 
            num--; 
        } 
        if(num > 0) ans.clear(); 
        return ans; 
    } 
};