LeetCode Course Schedule II

题目：

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.

click to show more hints.

Hints:

1. This problem is equivalent to finding the topological order in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
2. Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
3. Topological sort could also be done via BFS.

题意：

此题是在上一题的基础上，要求输出其中的某一个可能的拓扑排序数组，如果有多个，那么只需要输出一个即可。方法与上一题一样，只是如果碰到不存在拓扑排序的，要求输出空的数组[ ]。这里一开始LZ不太明白这个空的数组是什么意思？后来看了网上的一些案例，其实就是int[] a = new int[0];

public int[] findOrder(int numCourses,int[][] prerequisites)
	{
		//boolean canFind = false;
		int[] result = new int[numCourses];
		List<Set> posts = new ArrayList<Set>();
		Queue<Integer> queue = new LinkedList<Integer>();
		for(int i = 0; i < numCourses; i++)
		{
			posts.add(new HashSet<Integer>());
		}
		for(int i = 0; i < prerequisites.length; i++)
		{
			posts.get(prerequisites[i][1]).add(prerequisites[i][0]);
		}
		int[] preNums = new int[numCourses];
		for(int i = 0; i < numCourses; i++)
		{
			Set<Integer> set = posts.get(i);
			Iterator<Integer> it = set.iterator();
			while(it.hasNext())
			{
				preNums[it.next()]++;
			}
		}
		for(int i = 0; i < numCourses; i++)
		{
			if(preNums[i] == 0)
				queue.add(i);
		}
		if(queue.size() == 0)
			return new int[0];
		result = new int[numCourses];
		int count = 0;
		while(!queue.isEmpty())
		{
			int k = queue.poll();
			result[count] = k;
			Set<Integer> s = posts.get(k);
			Iterator<Integer> ite = s.iterator();
			while(ite.hasNext())
			{
				int q = ite.next();
				preNums[q] = preNums[q] - 1;
				if(preNums[q] == 0)
					queue.add(q);
			}
			count = count + 1;
		}
		if(count == numCourses)
			return result;
		else 
			return new int[0];
	}