标签：== cat exists targe [] xpl solution sum while
https://leetcode.com/problems/course-schedule-ii/#/description
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:
Hints:
Sol 1:
BFS.
import collections class Solution(object): def findOrder(self, numCourses, prerequisites): """ :type numCourses: int :type prerequisites: List[List[int]] :rtype: List[int] """ # bfs dic = {i: set() for i in range(numCourses)} neigh = collections.defaultdict(set) for i, j in prerequisites: dic[i].add(j) neigh[j].add(i) # queue stores the courses which have no prerequisites queue = collections.deque([i for i in dic if not dic[i]]) count, res = 0, [] while queue: # examine each node from queue node = queue.popleft() res.append(node) count += 1 # for each neighbors of the node for i in neigh[node]: # remove the prerequisites of this course i dic[i].remove(node) # if course i does not have any prerequisites, append it to queue if not dic[i]: queue.append(i) if count == numCourses: return res else: return []
Sol 2 :
DFS
import collections class Solution(object): def findOrder(self, numCourses, prerequisites): """ :type numCourses: int :type prerequisites: List[List[int]] :rtype: List[int] """ # dfs dic = collections.defaultdict(set) neigh = collections.defaultdict(set) for i, j in prerequisites: dic[i].add(j) neigh[j].add(i) # stack stores the courses which have no prerequisites stack = [i for i in range(numCourses) if not dic[i]] res = [] while stack: # examine each node from queue node = stack.pop() res.append(node) # for each neighbors of the node for i in neigh[node]: # remove the prerequisites of this course i dic[i].remove(node) # if course i does not have any prerequisites, append it to stack if not dic[i]: stack.append(i) dic.pop(node) if not dic: return res else: return []
标签：== cat exists targe [] xpl solution sum while
原文地址：http://www.cnblogs.com/prmlab/p/7229801.html
