Course Schedule II(topo)

There are a total of_n_courses you have to take, labeled from0ton-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 prerequisitepairs, 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.

分析

最后比较q(或者res)是否 == N, 不是比Indegree

这里的bfs的q可以用while和pop【0】

或者 for cur in q,就不用pop[0]

class Solution:
    def findOrder(self, numCourses, prerequisites):
        """
        :type numCourses: int
        :type prerequisites: List[List[int]]
        :rtype: List[int]
        """
        if not numCourses:
            return []
        graph = {i:[] for i in range(numCourses)}
        indegree = {i:0 for i in range(numCourses)}
        for p in prerequisites:
            graph[p[1]].append(p[0])
            indegree[p[0]]+=1
        q=[i for i in indegree if indegree[i]==0]

        for cur in q:
            for nn in graph[cur]:
                indegree[nn]-=1
                if indegree[nn] == 0:
                    q.append(nn)

        return q if len(q) == numCourses else []