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:

1. The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.

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

解题要点：

与上一题想法类似，不同的是，在queue里处理的每个元素都加进返回值里，如果返回值不足课程数量，说明满足不了上完全部的课，返回空，除此之外直接返回最终值。

class Solution(object):
    def findOrder(self, numCourses, prerequisites):
        """
        :type numCourses: int
        :type prerequisites: List[List[int]]
        :rtype: List[int]
        """
        res = []
        indegrees = [0] * numCourses
        mydict = collections.defaultdict(list)
        for i in range(len(prerequisites)):
            indegrees[prerequisites[i][0]] += 1
            mydict[prerequisites[i][1]] += prerequisites[i][0],
        
        queue = []
        for n in range(numCourses):
            if indegrees[n] == 0:
                queue.append(n)
        
        while queue: 
            p = queue.pop(0)
            res.append(p)       
            for e in mydict[p]:
                indegrees[e] -= 1
                if indegrees[e] == 0:
                    queue.append(e)
        if len(res) == numCourses:
            return res
        else:
            return []