Course Schedule

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, is it possible for 

you to finish all courses?

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 it is possible.

2, [[1,0],[0,1]]

There are a total of 2 courses to take. To take course 1 you should have finished 

course 0, and to take course 0 you should also have finished course 1. So it is impossible.

图有两种表示方法，邻接表和邻接矩阵。但这个题只要使用BFS简单处理即可。

找到所有没有先修课程的节点放入队列。数组中存放所有课程，如果课程是先修课程，

就+1。然后使用队列BFS遍历。队列中的课程有先修，则数组中对于的课程-1；如果

此时课程为0，则count+1，最后返回count==numCourses。

这个过程实际上是判断图中是否存在环。

public boolean canFinish(int numCourses, int[][] prerequisites) {
	if(prerequisites == null)
		throw new IllegalArgumentException("illegal prerequisites array");
	int len = prerequisites.length;
	if(numCourses == 0 || len == 0)
		return true;
	// counter for number of prerequisites
	int[] pCounter = new int[numCourses];
	for(int i=0; i<len; i++)
		pCounter[prerequisites[i][0]]++;
	//store courses that have no prerequisites
	LinkedList<Integer> queue = new LinkedList<Integer>();
	for(int i=0; i<numCourses; i++){
		if(pCounter[i]==0)
			queue.add(i);
	}
	// number of courses that have no prerequisites
	int numNoPre = queue.size();
	while(!queue.isEmpty()){
		int top = queue.remove();
		for(int i=0; i<len; i++){
			// if a course's prerequisite can be satisfied by a course in queue
			if(prerequisites[i][1]==top){
				pCounter[prerequisites[i][0]]--;
				if(pCounter[prerequisites[i][0]]==0){
					numNoPre++;
					queue.add(prerequisites[i][0]);
				}
			}
		}
	}
	return numNoPre == numCourses;
}