[leetcode] 210. Course Schedule II

There are a total of n courses you have to take labelled from 0 to n - 1.

Some courses may have prerequisites, for example, if prerequisites[i] = [ai, bi] this means you must take the course bi before the course ai.

Given the total number of courses numCourses and a list of the prerequisite pairs, return the ordering of courses you should take to finish all courses.

If there are many valid answers, return any of them. If it is impossible to finish all courses, return an empty array.

 

Example 1:

Input: numCourses = 2, prerequisites = [[1,0]] Output: [0,1] Explanation: 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].

 

Example 2:

Input: numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]] Output: [0,2,1,3] Explanation: 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].

 

Example 3:

Input: numCourses = 1, prerequisites = [] Output: [0]

 

Constraints:

• 1 <= numCourses <= 2000
• 0 <= prerequisites.length <= numCourses * (numCourses - 1)
• prerequisites[i].length == 2
• 0 <= ai, bi < numCourses
• ai != bi
• All the pairs [ai, bi] are distinct.

 

topological sort 를 사용해서 풀었다.

 

inDegree 가 0 인것부터 하나씩 간선을 없애나가면서 result list 를 만들어 나간다.

 

모든 course 를 끝낼 수 있는 valid answer case 에서는 result list 가 numCourses 와 같아야 한다.

 

java submissions 중에 33 프로를 이긴걸로 봐서 더 최적화가 가능할듯 싶지만.. 오늘은 여기까지만 해야겠다..

 

class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {

		Map<Integer, Node> graph = new HashMap<>();

		for (int i = 0; i < numCourses; i++) {
			graph.put(i, new Node(i));
		}

		for (int [] prerequisite : prerequisites) {
			Node previous = graph.computeIfAbsent(prerequisite[1], t -> new Node(prerequisite[1]));
			Node next = graph.computeIfAbsent(prerequisite[0], t -> new Node(prerequisite[0]));

			next.inDegrees += 1;
			previous.outNodes.add(next);
		}

		LinkedList<Node> noInDegreeList = new LinkedList<>();

		for (Node node : graph.values()) {
			if (node.inDegrees == 0) {
				noInDegreeList.add(node);
			}
		}

		List<Integer> res = new ArrayList<>();

		while (noInDegreeList.size() > 0) {
			Node node = noInDegreeList.removeFirst();
			res.add(node.val);

			for (Node child : node.outNodes) {
				child.inDegrees -= 1;

				if(child.inDegrees == 0) {
					noInDegreeList.add(child);
				}
			}
		}

		if (res.size() != numCourses) {
			return new int[0];
		}

		return res.stream().mapToInt(i -> i).toArray();
	}

	static class Node {
		int inDegrees;
		int val;
		List<Node> outNodes = new ArrayList<>();

		public Node(int val) {
			this.val = val;
		}
	}
}