Course Schedule II

Problem

There are a total of n courses you have to take, labeled from 0 to n - 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 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 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.

Solution

Topological

public class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        int[] res = new int[numCourses], degrees = new int[numCourses];
        for (int[] p : prerequisites) degrees[p[0]]++;
        Queue<Integer> q = new LinkedList<>();
        for (int i = 0; i < numCourses; i++) {
            if (degrees[i] == 0) q.offer(i);
        }
        int index = 0;
        while (!q.isEmpty()) {
            int course = q.poll();
            res[index++] = course;
            for (int[] p : prerequisites) {
                if (p[1] == course) {
                    degrees[p[0]]--;
                    if (degrees[p[0]] == 0) q.offer(p[0]);
                }
            } 
        }
        return index == numCourses ? res : new int[0];
    }
}

Analysis

A typical topological solution
We first loop through the given prerequisites and update degree of each class
Then we loop through the degrees and offer class whose degree is 0
Then we use BFS to update our res and loop through the prerequisites again to update degrees
We offer new course to the queue if degrees of that course becomes 0 after updating
We have index to record the index of added course and we use it to check whether we can complete all courses