-
Notifications
You must be signed in to change notification settings - Fork 0
/
Copy path29. Course Schedule.cpp
71 lines (50 loc) · 1.88 KB
/
29. Course Schedule.cpp
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
/*
There are a total of numCourses courses you have to take, labeled from 0 to numCourses-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: numCourses = 2, prerequisites = [[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: numCourses = 2, prerequisites = [[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.
Constraints:
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.
1 <= numCourses <= 10^5
*/
class Solution {
public:
bool isCyclic(vector<int> adj[], vector<int>& visited, int v)
{
if(visited[v]==1)
return true;
if(visited[v]==0)
visited[v]=1;
for(auto edge:adj[v])
{
if(isCyclic(adj,visited,edge))
return true;
}
visited[v]=2;
return false;
}
bool canFinish(int numCourses, vector<vector<int>>& prerequisites)
{
vector<int>adj[numCourses];
for(auto &x:prerequisites)
adj[x[1]].push_back(x[0]);
vector<int>visited(numCourses,0);
for(int i=0;i<numCourses;i++)
{
if(isCyclic(adj,visited,i))
return false;
}
return true;
}
};