【LeetCode】210. Course Schedule II

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].

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.

 

参考Course Schedule，不断删除出度为0的点，该点即下一个选修的课程。

class Solution {
public:
    vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<int> order;
        vector<int> outd(numCourses, 0);
        vector<bool> del(numCourses, false);
        unordered_map<int, vector<int> > graph;
        // construct reverse neighborhood graph
        // graph is to decrease the out-degree of a set of vertices, 
        // when a certain vertice is deleted
        for(int i = 0; i < prerequisites.size(); i ++)
        {
            outd[prerequisites[i].first] ++;
            graph[prerequisites[i].second].push_back(prerequisites[i].first);
        }
        int count = 0;
        while(count < numCourses)
        {
            int i;
            for(i = 0; i < numCourses; i ++)
            {
                if(outd[i] == 0 && del[i] == false)
                    break;
            }
            if(i < numCourses)
            {
                del[i] = true;  // delete
                order.push_back(i);
                for(int j = 0; j < graph[i].size(); j ++)
                {// decrease the degree of vertices that links to vertice_i
                    outd[graph[i][j]] --;
                }
                count ++;
            }
            else
            {// no vertice with 0-degree
                vector<int> noorder;
                return noorder;
            }
        }
        return order;
    }
};