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].

与上一题不同的是，输出有向图的顺序。需要用一个数组存储队列弹出的元素。

public int[] findOrder(int numCourses, int[][] prerequisites) {
			int[] order=new int[numCourses];
	int len=prerequisites.length;
	if(prerequisites==null)
		throw new IllegalArgumentException("illegal prerequisites array");
	if(len==0){
		for(int i=0;i<numCourses;i++)
			order[i]=i;
		return order;
	}
	for(int i=0; i<len; i++)
		order[prerequisites[i][0]]++;
	Queue<Integer> queue = new LinkedList<Integer>();
	for(int i=0; i<numCourses; i++){
		if(order[i]==0)
			queue.add(i);
	}
	int numNoPre = queue.size();
	int[] result=new int[numCourses];
	int j=0;
	while(!queue.isEmpty()){
		int top = queue.poll();
		result[j++]=top;
		for(int i=0; i<len; i++){
			if(prerequisites[i][1]==top){
				order[prerequisites[i][0]]--;
				if(order[prerequisites[i][0]]==0){
					numNoPre++;
					queue.add(prerequisites[i][0]);
				}
			}
		}
	}
	if(numNoPre==numCourses)
		return result;
	else
		return new int[0];
}