-
Notifications
You must be signed in to change notification settings - Fork 1
/
Copy path207.course-schedule.py
79 lines (77 loc) · 2.26 KB
/
207.course-schedule.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
#
# @lc app=leetcode id=207 lang=python
#
# [207] Course Schedule
#
# https://leetcode.com/problems/course-schedule/description/
#
# algorithms
# Medium (37.30%)
# Total Accepted: 204.4K
# Total Submissions: 546.7K
# Testcase Example: '2\n[[1,0]]'
#
# 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.
#
#
#
class Solution(object):
def canFinish(self, numCourses, prerequisites):
"""
:type numCourses: int
:type prerequisites: List[List[int]]
:rtype: bool
"""
self.visited = [0 for i in range(numCourses)]
self.courses = [[0 for i in range(numCourses)] for j in range(numCourses)]
self.flag = True
for prerequisite in prerequisites:
i, j = prerequisite
self.courses[i][j] = 1
for i in range(numCourses):
self.dfs(i)
return self.flag
def dfs(self, s):
if self.visited[s] == -1:
return
self.visited[s] = 1
for i in range(len(self.visited)):
if self.courses[s][i]:
if not self.visited[i]:
self.dfs(i)
elif self.visited[i] == 1:
self.flag = False
return
self.visited[s] = -1