# 210. Course Schedule II

<https://leetcode.com/problems/course-schedule-ii/>

## solution

```python
import collections

class Solution:
    def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]:
        courses = collections.defaultdict(set)
        pres = collections.defaultdict(set)

        for (x, y) in prerequisites:
            courses[y].add(x)  # 上完这个课之后可以上的
            pres[x].add(y)  # 这个课程的所有先修课程

        no_pre = [i for i in range(numCourses) if not pres[i]]

        res = []
        count = 0
        while no_pre:
            take_course = no_pre.pop()
            count += 1
            res.append(take_course)
            for i in courses[take_course]:
                pres[i].remove(take_course)
                if not pres[i]:
                    no_pre.append(i)

        return res if count == numCourses else []
```

时间复杂度：O(V + E)\
空间复杂度：O(V + E)


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