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:
- 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.
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
}
}