[LeetCode] Course Schedule II Problem Solving report, leetcodeschedule
[Question]
There are a total of n courses you have to take, labeled from0
Ton - 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, return the ordering of courses you shoshould take to finish all courses.
There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.
For example:
2, [[1,0]]
There are a total of 2 courses to take. To take course 1 you shoshould have finished course 0. So the correct course order is[0,1]
4, [[1,0],[2,0],[3,1],[3,2]]
There are a total of 4 courses to take. to take course 3 you shoshould have finished both courses 1 and 2. both courses 1 and 2 shoshould be taken after you finished course 0. so one correct course order is[0,1,2,3]
. Another correct ordering is[0,2,1,3]
.
[Resolution]
Compared with the [LeetCode] Course Schedule problem solving report, the returned results are a group of sorting results rather than whether they can be sorted.
As long as any valid Topology Sorting result is returned, you only need to record the accessed nodes in the BFS process.
[Java code]
public class Solution { public int[] findOrder(int numCourses, int[][] prerequisites) { List<Set<Integer>> adjLists = new ArrayList<Set<Integer>>(); for (int i = 0; i < numCourses; i++) { adjLists.add(new HashSet<Integer>()); } for (int i = 0; i < prerequisites.length; i++) { adjLists.get(prerequisites[i][1]).add(prerequisites[i][0]); } int[] indegrees = new int[numCourses]; for (int i = 0; i < numCourses; i++) { for (int x : adjLists.get(i)) { indegrees[x]++; } } Queue<Integer> queue = new LinkedList<Integer>(); for (int i = 0; i < numCourses; i++) { if (indegrees[i] == 0) { queue.offer(i); } } int[] res = new int[numCourses]; int count = 0; while (!queue.isEmpty()) { int cur = queue.poll(); for (int x : adjLists.get(cur)) { indegrees[x]--; if (indegrees[x] == 0) { queue.offer(x); } } res[count++] = cur; } if (count == numCourses) return res; return new int[0]; }}