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.

For example:

1
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]

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.

要求返回一个图的拓扑排序结果，在有向图存在环路的时候返回空集。拓扑排序可以用点度和队列结合的办法，也可以用DFS的办法。因为输入数据是邻接表，对边的处理比较慢，所以我用DFS来实现拓扑排序，同时判断是否有环存在。

这个题有个条件就是节点编号是在0-n-1之间的，所以记录一个节点的状态可以用数组，没有查询的开销。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
class Solution {
    vector<int> visited;
    vector<int> inPath;
public:
    vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<vector<int>> nextCourses(numCourses), preReq(numCourses);
        for(auto i : prerequisites){
            preReq[i.first].push_back(i.second);
            nextCourses[i.second].push_back(i.first);
        }
        visited = vector<int>(numCourses, 0);
        inPath = vector<int>(numCourses, 0);
        vector<int> ans;
        for(int i = 0; i < numCourses; i++){
            if(!visited[i] && preReq[i].empty()){
                // Begin node
                vector<int> t, path;
                if(DFS(nextCourses, preReq, i, path, t))
                    ans.insert(ans.end(), t.begin(), t.end());
                else
                    return vector<int>();
            }
        }
        if(ans.size() != numCourses)
            return vector<int>();
        else
            return ans;
    }
    
    bool DFS(vector<vector<int>> &nextCourses, vector<vector<int>> &preReq, int node, vector<int> &path, vector<int> &ans){
        if(inPath[node]) {
            ans.clear();
            return false;
        }
        if(visited[node]) return true;
        for(auto i : preReq[node]){
            if(!visited[i])
                return true;
        }
        
        path.push_back(node);
        ans.push_back(node);
        inPath[node] = visited[node] = 1;
        
        for(auto i : nextCourses[node]){
            if(!DFS(nextCourses, preReq, i, path, ans)) 
                return false;
        }
        path.pop_back();
        inPath[node] = 0;
        return true;
    }
};