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] .

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.

难道: medium

题目:
课程编号为0到n-1的n门课程需要我们去学习。

学习某些课程需要先学习其预备课程，例如学习课程0就要先学习课程1, 用如下形式表示这种关系[0, 1].

给定全部课程数，以及这种课程与课程之前的关系，返回可以学习完所有课程的顺序。
完成所有课程的正确顺序有许多种，你只需要返回其中正确的一条学习路径。如果没有可以学完所有课程的顺序，则返回空数组。

注意：
前置课程是由一些边组成的图不是毗邻的矩阵。
你可以假定前置课程这种关系表示没有重复的输入。

思路：
BFS, 并保留遍历过程中入度为0的结点。
拓扑图遍历
1.找出一个出度为0的结点。
2.移除该出度为0结点的所有边，然后执行1.

Runtime: 8 ms, faster than 72.40% of Java online submissions for Course Schedule II.

class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        // (Array -> List) to store graph
        List<Integer>[] nodes = new ArrayList[numCourses];
        for (int i = 0; i < numCourses; i++) {
            nodes[i] = new ArrayList<Integer>();
        }
        
        // count in degree
        int[] inDegree = new int[numCourses];
        for (int i = 0; i < prerequisites.length; i++) {
            (nodes[prerequisites[i][0]]).add(prerequisites[i][1]);
            inDegree[prerequisites[i][1]] += 1;
        }
        
        // count zero in degree
        int zeroInDegreeCount = 0;
        List<Integer> zeroInDegreeList = new ArrayList<>();
        for (int i = 0; i < inDegree.length; i++) {
            if (inDegree[i] <= 0) {
                zeroInDegreeList.add(i);
                zeroInDegreeCount++;
            }
        }
        
        // bfs
        for (int i = 0; i < zeroInDegreeList.size(); i++) {
            for (Integer node : nodes[zeroInDegreeList.get(i)]) {
                if (--inDegree[node] <= 0) {
                    zeroInDegreeList.add(node);
                    zeroInDegreeCount++;
                }
            }
        }
        
        if (zeroInDegreeCount == numCourses) {
            int[] result = new int[numCourses];
            for (int i = 0; i < numCourses; i++) {
                result[numCourses - 1 - i] = zeroInDegreeList.get(i);
            }
            return result;
        } 
        
        return new int[0];
    }
}