LeetCode #210 - Course Schedule II-优快云博客

本文介绍了一种基于图的课程排序算法，用于确定完成所有课程的一种有效顺序。通过构建课程依赖图并利用拓扑排序原理，该算法能有效地解决课程先修顺序问题。

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题目描述：

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.

和Course Schedule一样的思路，只是需要求出一个可行的选课顺序，其实就是在每次访问节点的同时，把节点加入最终结果即可。

class Solution {
public:
    vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<int> empty;
        vector<vector<int>> graph(numCourses,empty);
        vector<int> indegree(numCourses,0);
        queue<int> start;
        vector<int> order;
        for(int i=0;i<prerequisites.size();i++)
        {
            graph[prerequisites[i].second].push_back(prerequisites[i].first);
            indegree[prerequisites[i].first]++;
        }
        for(int i=0;i<numCourses;i++)
        {
            if(indegree[i]==0) 
            {
                start.push(i);
                order.push_back(i);
            }
        }
        
        while(!start.empty())
        {
            int i=start.front();
            for(int j=0;j<graph[i].size();j++)
            {
                indegree[graph[i][j]]--;
                if(indegree[graph[i][j]]==0) 
                {
                    start.push(graph[i][j]);
                    order.push_back(graph[i][j]);
                }
            }
            start.pop();
        }
        if(order.size()==numCourses) return order;
        else return empty;
    }
};