[Leetcode]Course Schedule II

Course Schedule II My Submissions Question
Total Accepted: 16442 Total Submissions: 80267 Difficulty: Medium
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.

Hints:
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.
Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
Topological sort could also be done via BFS.
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刚开始以为肯定是能正确输出，一次AC。提交了下发现如果有环的话就不可以了。在结尾出加了个判断，只有res内的数量和课程数目一样时候才有拓扑序列。
其他情况都是没有的。
再提交就对了！而且居然能打败97%的C++答案。。。

class Solution {
public:
    vector<int> findOrder(int n, vectorint, int>>& pre) {
        vector<vector<int> > graph(n);
        vector<int> degree(n,0);
        vector<int> res;
        for(auto p : pre){
            graph[p.second].push_back(p.first);
            ++ degree[p.first];
        }
        queue<int> q;
        for(int i = 0;i <= n - 1;++ i){
            if(degree[i] == 0){
                q.push(i);
            }
        }
        while(!q.empty()){
            int course = q.front();
            q.pop();
            res.push_back(course);
            for(auto neigh : graph[course]){
                --degree[neigh];
                if(degree[neigh] == 0){
                    q.push(neigh);
                }
            }
        }
        if(res.size() != n){
            res.clear();
            return res;
        }
        return res;
    }
};