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].

在Course Schedule 1的基础上输出一个可以完成所有课程的序列。只需要在递归完之后记录节点就可以了。代码如下：

public class Solution {
    boolean[] onStack;
    boolean[] isVisited;
    LinkedList<Integer> list;
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        int[] result = new int[numCourses];
        list = new LinkedList<Integer>();
        List<Integer>[] graph = new List[numCourses];
        onStack = new boolean[numCourses];
        isVisited = new boolean[numCourses];
        for(int i = 0; i < prerequisites.length; i++) {
            int key = prerequisites[i][1];
            int value = prerequisites[i][0];
            if(graph[key] == null) graph[key] = new ArrayList<Integer>();
            graph[key].add(value);
        }
        for(int i = 0; i < numCourses; i++) {
            if(isVisited[i] == false && hasCycle(i, graph) == true)
                return new int[0];
        }
        for(int i = 0; i < list.size(); i++) {
            result[i] = list.get(i);
        }
        return result;
    }
    
    public boolean hasCycle(int i, List<Integer>[] graph) {
        boolean cycle = false;
        onStack[i] = true;
        isVisited[i] = true;
        if(graph[i] != null) {
            for(int c : graph[i]) {
                if(isVisited[c] == false) {
                    cycle = hasCycle(c, graph);
                    if(cycle == true)
                        break;
                } else if(onStack[c] == true) {
                    cycle = true;
                    break;
                }
            }
        }
        onStack[i] = false;
        list.addFirst(i);
        return cycle;
    }
}