LeetCode-in-All
207. Course Schedule
Medium
There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course bi first if you want to take course ai.
- For example, the pair
[0, 1], indicates that to take course0you have to first take course1.
Return true if you can finish all courses. Otherwise, return false.
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:
1 <= numCourses <= 20000 <= prerequisites.length <= 5000prerequisites[i].length == 20 <= ai, bi < numCourses- All the pairs prerequisites[i] are unique.
Solution
type State int
const (
Unvisited State = iota
Visiting
Visited
)
func canFinish(numCourses int, prerequisites [][]int) bool {
visited := make([]State, numCourses)
graph := make([][]int, numCourses)
for _, dep := range prerequisites {
graph[dep[1]] = append(graph[dep[1]], dep[0])
}
for i := 0; i < numCourses; i++ {
if visited[i] == Unvisited {
if !dfs(i, visited, graph) {
return false
}
}
}
return true
}
func dfs(start int, visited []State, graph [][]int) bool {
if visited[start] == Visiting {
return false
}
if visited[start] == Visited {
return true
}
visited[start] = Visiting
for _, next := range graph[start] {
if !dfs(next, visited, graph) {
return false
}
}
visited[start] = Visited
return true
}