难度: Medium
原题连接
内容描述
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, is it possible for you to finish all courses?
Example 1:
Input: 2, [[1,0]]
Output: true
Explanation: There are a total of 2 courses to take.
To take course 1 you should have finished course 0. So it is possible.
Example 2:
Input: 2, [[1,0],[0,1]]
Output: false
Explanation: There are a total of 2 courses to take.
To take course 1 you should have finished course 0, and to take course 0 you should
also have finished course 1. So it is impossible.
Note:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
You may assume that there are no duplicate edges in the input prerequisites.
思路 1 - 时间复杂度: O(V+E)- 空间复杂度: O(N)******
Topological sorting for Directed Acyclic Graph (DAG) is a linear ordering of vertices such that for every directed edge uv, vertex u comes before v in the ordering. Topological Sorting for a graph is not possible if the graph is not a DAG.
pre-requisite问题，只需要判断最终的topological结果长度与courses数目是否相等即可
DFS 和 BFS都可以用来拓扑排序。
beats 95.38%
class Solution(object):
def canFinish(self, numCourses, prerequisites):
"""
:type numCourses: int
:type prerequisites: List[List[int]]
:rtype: bool
"""
graph = collections.defaultdict(list)
indegrees = [0] * numCourses
for course, pre in prerequisites:
graph[pre].append(course)
indegrees[course] += 1
return self.topologicalSort(graph, indegrees) == numCourses
def topologicalSort(self, graph, indegrees):
count = 0
queue = []
for i in range(len(indegrees)):
if indegrees[i] == 0:
queue.append(i)
while queue:
course = queue.pop()
count += 1
for i in graph[course]:
indegrees[i] -= 1
if indegrees[i] == 0:
queue.append(i)
return count