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

1. This problem is equivalent to finding the topological order in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
2. Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
3. Topological sort could also be done via BFS.

這題是之前那道 Course Schedule 課程清單的擴充套件，那道題只讓我們判斷是否能完成所有課程，即檢測有向圖中是否有環，而這道題我們得找出要上的課程的順序，即有向圖的拓撲排序，這樣一來，難度就增加了，但是由於我們有之前那道的基礎，而此題正是基於之前解法的基礎上稍加修改，我們從queue中每取出一個數組就將其存在結果中，最終若有向圖中有環，則結果中元素的個數不等於總課程數，那我們將結果清空即可。程式碼如下：

class Solution {
public:
    vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<int> res;
        vector<vector<int> > graph(numCourses, vector<int>(0));
        vector<int> in(numCourses, 0);
        for (auto &a : prerequisites) {
            graph[a.second].push_back(a.first);
            ++in[a.first];
        }
        queue<int> q;
        for (int i = 0; i < numCourses; ++i) {
            if (in[i] == 0) q.push(i);
        }
        while (!q.empty()) {
            int t = q.front();
            res.push_back(t);
            q.pop();
            for (auto &a : graph[t]) {
                --in[a];
                if (in[a] == 0) q.push(a);
            }
        }
        if (res.size() != numCourses) res.clear();
        return res;
    }
};

類似題目：