LeetCode 210. Course Schedule II

Posted 2021-01-01 sylvanyang

题目链接：https://leetcode.com/problems/course-schedule-ii/submissions/1

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.

这道题连同Course Schedule I 一起，都是Topology Sorting的应用。只要实现了拓扑排序，那么题目基本上完全没有任何变形。代码的区别也仅在于return的对象不同而已。代码如下：

class Solution(object):
    def findOrder(self, numCourses, prerequisites):
        """
        :type numCourses: int
        :type prerequisites: List[List[int]]
        :rtype: List[int]
        """
        graph = {}
        inDegree = collections.defaultdict(int)
        for t, s in prerequisites:
            if s not in graph:
                graph[s] = [t]
            else:
                graph[s].append(t)
            inDegree[t] += 1
        path = []
        for x in range(numCourses):
            if x not in inDegree:
                inDegree[x] = 0
        
        #start Topology sorting
        while True:
            startLength = len(path)
            for k in inDegree.keys():
                if inDegree[k] == 0:
                    path.append(k)
                    del inDegree[k]
                    #inDegree[k] = -1
                    if k in graph:
                        for item in graph[k]:
                            inDegree[item] -= 1
            if startLength == len(path):
                break
        return path if len(path) == numCourses else []