> For the complete documentation index, see [llms.txt](https://hao-fu-1.gitbook.io/oj/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://hao-fu-1.gitbook.io/oj/search/topological-sort/course-schedule-ii.md).

# 210. Course Schedule II

https\://leetcode.com/problems/course-schedule-ii/description/

> 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.
>
> You may assume that there are no duplicate edges in the input prerequisites.

## Thoughts

给一系列pair代表上第一个元素代表的课前必须先上第二个元素，找出一个合理的上课顺序能上完所有课程，如果不存在则返回空。有先后顺序的图，拓扑排序。依次BFS遍历入度为0的点并把相应的边删去，随着遍历依次把入度为0输出。

## Code

```java
/*
 * @lc app=leetcode id=210 lang=cpp
 *
 * [210] Course Schedule II
 */

// @lc code=start
class Solution {
public:
    vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) {
        vector<int> in(numCourses, 0);
        vector<unordered_set<int>> edges(numCourses, unordered_set<int>());
        for (const auto &p : prerequisites) {
            edges[p[1]].insert(p[0]);
            ++in[p[0]];
        } 
        queue<int> q;
        for (int i = 0; i < numCourses; ++i) {
            if (in[i] == 0) q.push(i);
        }
        vector<int> res;
        while (!q.empty()) {
            auto t = q.front(); q.pop();
            res.push_back(t);
            for (auto e : edges[t]) {
                --in[e];
                if (in[e] == 0) q.push(e);
            }
        }
        return res.size() == numCourses ? res : vector<int>();
    }
};
// @lc code=end


```

## Analysis

时间复杂度O(N), N为边数.