LeetCode 210 Course Schedule II - Medium

210. Course Schedule II — Medium

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Problem

210. Course Schedule II

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.

Example 1:

Input: 2, [[1,0]] Output: [0,1] Explanation: 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] . Example 2:

Input: 4, [[1,0],[2,0],[3,1],[3,2]] Output: [0,1,2,3] or [0,2,1,3] Explanation: 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.

Solution

### Topological Sorting:
 
class Solution:
    """
    @param: numCourses: a total of n courses
    @param: prerequisites: a list of prerequisite pairs
    @return: the course order
    """
    def findOrder(self, numCourses, prerequisites):
        # write your code here
        ### 建立graph
        ### 邻居list（此处的邻居为prerequisites）
        edges = {i : [] for i in range(numCourses) } 
        ### 入度
        degrees = [0 for i in range(numCourses)]
        
        for i, j in prerequisites:
            edges[j].append(i)
            degrees[i] += 1
            
        queue = collections.deque([])
        order = []
        
        ### 把所有入度为0的node放进queue
        for i in range(numCourses):
            if degrees[i] == 0:
                queue.append(i)
 
        while queue:
            node = queue.popleft()
            order.append(node)
            # 将每条邻边删去，如果入度降为 0，再加入队列
            for x in edges[node]:
                degrees[x] -= 1
                if degrees[x] == 0:
                    queue.append(x)
 
        if len(order) == numCourses:
            return order
        return []