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.
Solutions
拓扑排序-BFS
从出入度的角度来计算，复杂度是 $O(n^2)$：
class Solution(object):
def canFinish(self, numCourses, prerequisites):
"""
:type numCourses: int
:type prerequisites: List[List[int]]
:rtype: bool
"""
graph = collections.defaultdict(list)
indegrees = collections.defaultdict(int)
for u, v in prerequisites:
graph[v].append(u)
indegrees[u] += 1
for i in range(numCourses):
zero_degree = False
for j in range(numCourses):
if indegrees[j] == 0:
zero_degree = True
break
if not zero_degree:
return False
indegrees[j] = -1
for node in graph[j]:
indegrees[node] -= 1
return True
# Runtime: 324 ms, faster than 15.97% of Python online submissions for Course Schedule.
# Memory Usage: 13.4 MB, less than 76.27% of Python online submissions for Course Schedule.
拓扑排序-DFS
class Solution(object):
def canFinish(self, numCourses, prerequisites):
"""
:type numCourses: int
:type prerequisites: List[List[int]]
:rtype: bool
"""
graph = collections.defaultdict(list)
for u, v in prerequisites:
graph[u].append(v)
# 0 = Unkown, 1 = visiting, 2 = visited
visited = [0] * numCourses
for i in range(numCourses):
if not self.dfs(graph, visited, i):
return False
return True
def dfs(self, graph, visited, i):
if visited[i] == 1: return False
if visited[i] == 2: return True
visited[i] = 1
for j in graph[i]:
if not self.dfs(graph, visited, j):
return False
visited[i] = 2
return True
# Runtime: 68 ms, faster than 99.28% of Python online submissions for Course Schedule.
# Memory Usage: 14.5 MB, less than 49.15% of Python online submissions for Course Schedule.