210. Course Schedule II【M】【7】

最新推荐文章于 2020-12-01 10:30:12 发布

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#图 #leetcode题解 #leetcode #python #拓扑排序

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本文介绍了一种基于拓扑排序解决课程先修顺序问题的算法。该算法适用于存在课程依赖关系的场景，通过构建图并利用拓扑排序求解课程的合理学习顺序。

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.

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直接用拓扑排序就行了。不过有个问题是，他给的pair是反的。。。

class Solution(object):
    def findOrder(self, numCourses, prerequisites):

        n = numCourses
        p = prerequisites
        v = [0] * n
        dic = {}
        indegree = [0] * n
        res = []

        for i in p:
            dic[i[1]] = dic.get(i[1],[]) + [i[0]]
            indegree[i[0]] += 1

        q = []
        for i in xrange(len(indegree)):
            if indegree[i] == 0:
                q += i,
        #print indegree,q
        while q != []:
            node = q.pop()
            res += node,
            for i,j in p:
                if j == node:
                    indegree[i] -= 1
                    if indegree[i] == 0:
                        q += i,
        #print res,indegree
        if sum(indegree)  == 0:
            return res
        return []