Course Schedule II

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].


/*
The problem reduces to find a topological sort order of the courses. If a node has incoming edges, it has prerequisites. Therefore, the first few in the order must be those with no prerequisites, i.e. no incoming edges. Notice, if there is a cycle in the graph, it means it is impossible to finish all the courses.

time:O(n), space:O(n)
*/

public class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        Map<Integer, Integer> inMap = countInlinks(prerequisites);
        Map<Integer, ArrayList<Integer>> outMap =                                                 initNeighbors(prerequisites);
        return findOrderByBFS(numCourses, inMap, outMap);
    }
    public Map<Integer, Integer> countInlinks(int[][] pre) {
        Map<Integer, Integer> map = new HashMap<>();
        for (int[] pair : pre) {
            if (!map.containsKey(pair[0])) {
                map.put(pair[0], 1);
            } else {
                map.put(pair[0], map.get(pair[0]) + 1);
            }
        }
        return map;
    }
    public Map<Integer, ArrayList<Integer>> initNeighbors(int[][]                                                             pre) {
        Map<Integer, ArrayList<Integer>> map = new HashMap<>();
        for (int[] pair : pre) {
            if (!map.containsKey(pair[1])) {
                ArrayList<Integer> list = new ArrayList<>();
                list.add(pair[0]);
                map.put(pair[1], list);
            } else {
                map.get(pair[1]).add(pair[0]);
            }
        }
        return map;
    }
    public int[] findOrderByBFS(int numCourses, Map<Integer,                   Integer> inMap, Map<Integer, ArrayList<Integer>> outMap) {
        int[] ans = new int[numCourses];
        Queue<Integer> queue = new LinkedList<>();
        int index = 0;
        for (int course = 0; course < numCourses; course++) {
            if (!inMap.containsKey(course)) {
                ans[index++] = course;
                queue.offer(course);
            }
        }
        while (!queue.isEmpty()) {
            int course = queue.poll();
            if (!outMap.containsKey(course)) {
                continue;
            }
            for (Integer i : outMap.get(course)) {
             inMap.put(i, inMap.get(i) - 1);
                if (inMap.get(i) == 0) {
                    ans[index++] = i;
                    queue.offer(i);
                }
            }
        }
        return index == numCourses ? ans : new int[0];
    }
}