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].


这题是Course Schedule的升级版本，在判断DAG的基础上要输出拓扑排序的结果。思路和Course Schedule一样，只是需要加一个构造输出结果的过程。

代码：

class Solution {
public:
    vector<int> findOrder(int numCourses, vector<pair<int, int> >& prerequisites) {
        vector<int> ret;
        vector<int> notpossible;
        if(numCourses == 0) return ret;

        vector<int> indeg(numCourses, 0);
        vector<vector<int> > edges(numCourses, vector<int>());
        vector<bool> visited(numCourses, false);
        int count = 0, nedge = prerequisites.size();

        for(int i = 0; i < nedge; ++i){
            ++indeg[prerequisites[i].first];
            edges[prerequisites[i].second].push_back(prerequisites[i].first);
        }

        while(count < numCourses){
            int i = 0;
            for(; i < numCourses; ++i) if(!visited[i] && indeg[i] == 0) break;
            if(i == numCourses) return notpossible;
            visited[i] = true;
            ++count;
            ret.push_back(i);
            int nout = edges[i].size();
            for(int j = 0; j < nout; ++j) --indeg[edges[i][j]];
        }
        return ret;
    }
};