[LeetCode206] Course Schedule II

There are a total ofncourses you have to take, labeled from0ton – 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 prerequisitepairs, return the ordering of courses you should 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 should 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 should have finished both courses 1 and 2. Both courses 1 and 2 should 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].

Note:The input prerequisites is a graph represented bya list of edges, not adjacency matrices. Read more abouthow a graph is represented.

solution: similar to "course schedule" need to record the path.

public int[] findOrder(int numCourses, int[][] prerequisites) {List<HashSet<Integer>> paths = new ArrayList<HashSet<Integer>>();for(int i=0; i<numCourses;i++){paths.add(new HashSet<Integer>());}//initialize graph — adjeced list graphfor(int i=0;i<prerequisites.length;i++){paths.get(prerequisites[i][1]).add(prerequisites[i][0]);}//calculate each node degreeint[] degrees = new int[numCourses];for(int i=0;i<numCourses;i++){for(int j: paths.get(i)){degrees[j]++;}}//store node which degree is 0Queue<Integer> road = new LinkedList<>();for(int i=0;i<numCourses;i++){if(degrees[i]==0) road.offer(i);}int[] result = new int[numCourses];int count = 0;while(!road.isEmpty()){int cur = road.poll();for(int i: paths.get(cur)){degrees[i]–;if(degrees[i]==0){road.offer(i);}}result[count++] = cur;}if(count == numCourses) return result;return new int[0];}

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