Leetcode** 210. Course Schedule II

Description: There are a total of n courses you have to take labelled from 0 to n - 1.

Some courses may have prerequisites, for example, if prerequisites[i] = [ai, bi] this means you must take the course bi before the course ai.

Given the total number of courses numCourses and a list of the prerequisite pairs, return the ordering of courses you should take to finish all courses.

If there are many valid answers, return any of them. If it is impossible to finish all courses, return an empty array.

Link: 210. Course Schedule II

Examples:

Example 1:
Input: numCourses = 2, prerequisites = [[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: numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]]
Output: [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].

Example 3:
Input: numCourses = 1, prerequisites = []
Output: [0]

思路: 这道题是207的升级版，需要在可以成功安排所有课程的情况下，返回任意一种安排方式。因为不是返回所有可能的结果，就变得简单很多。只需要在207的基础上，添加一条路径。

class Solution(object):
    def findOrder(self, numCourses, prerequisites):
        """
        :type numCourses: int
        :type prerequisites: List[List[int]]
        :rtype: List[int]
        """
        self.pre = collections.defaultdict(list)
        for p in prerequisites:
            self.pre[p[0]].append(p[1])
        self.visited = [0]*numCourses
        self.path = []
        for i in range(numCourses):
            if not self.dfs(i):
                return []
        return self.path
    
    def dfs(self, i):
        if self.visited[i] == 1: return True
        if self.visited[i] == 2: return False
        self.visited[i] = 2
        for b in self.pre[i]:
            if not self.dfs(b): return False
        self.visited[i] = 1
        self.path.append(i)
        return True

日期: 2021-04-20  时光不辜负