48.Course Schedule（课程安排）

Level：

  Medium

题目描述：

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?

Example 1:

Input: 2, [[1,0]] 
Output: true
Explanation: There are a total of 2 courses to take. 
             To take course 1 you should have finished course 0. So it is possible.

Example 2:

Input: 2, [[1,0],[0,1]]
Output: false
Explanation: 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.

Note:

1. The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
2. You may assume that there are no duplicate edges in the input prerequisites.

思路分析：

  这题是拓扑排序一道应用，首先，我们应该统计每个点的入度，然后根据拓扑排序的规则，先删去入度为0的点，直到最后点的个数为0，则是正确的排课。

代码：

public class Solution{
    public boolean canFinish(int coursenum,int[][]prerequisites){
        int []indegree=new int [coursenum];//记录每门课的入度。
        for(int []pair:prerequisites){
            indegree[pair[0]]++; //统计每门课的入度
        }
        Queue<Integer>q=new LinkedList<>(); //存放入度为0的课程，准备删除
        for(int i=0;i<coursenum;i++){
            if(indegree[i]==0)
                q.offer(i);
        }
        while(!q.isEmpty()){
            int key=q.poll();//删除一个入度为0的课程
            coursenum--;//课程数减1
            for(int[] pair:prerequisites){
                if(pair[1]==key){//因为删除了节点，那么和这个节点相连的节点入度减一。
                    indegree[pair[0]]--;
                if(indegree[pair[0]]==0)//如果减一后，这个节点的入度为0，则加入删除队列
                    q.offer(pair[0]);
                } 
            }
        }
        return coursenum==0; //如果最终都被删除，则满足排课顺序
    }
}