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]

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.

Example 1:

Input: 2, [[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: 4, [[1,0],[2,0],[3,1],[3,2]]
Output: [0,1,2,3] or [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] .

思路：跟Course Schedule I 很类似，只需要加一个int array来记录输出的点的顺序就可以了。注意一点：empty array 就是 array has no element,  with 0 length；new int[0];

class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        // [0, 1] -> 1 -> 0;
        HashMap<Integer, List<Integer>> graph = new HashMap<>();
        int[] indegree = new int[numCourses];
        for(int[] prerequist: prerequisites) {
            int a = prerequist[0];
            int b = prerequist[1];
            // b -> a;
            graph.putIfAbsent(b, new ArrayList<Integer>());
            graph.get(b).add(a);
            indegree[a]++;
        }
        
        Queue<Integer> queue = new LinkedList<>();
        HashSet<Integer> visited = new HashSet<>();
        for(int i = 0; i < numCourses; i++) {
            if(indegree[i] == 0) {
                queue.offer(i);
            }
        }
        
        int[] res = new int[numCourses];
        int index = 0;
        while(!queue.isEmpty()) {
            Integer node = queue.poll();
            if(!visited.contains(node)) {
                visited.add(node);
                res[index++] = node;
                if(graph.get(node) != null) {
                   for(Integer neighbor: graph.get(node)) {
                        indegree[neighbor]--;
                        if(indegree[neighbor] == 0) {
                            queue.offer(neighbor);
                        }
                    } 
                }
            }
        }
        
        return visited.size() == numCourses ? res : new int[]{};
    }
}