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.

Note:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.

这题主要用到的是拓扑排序。每次计算入度为0的点，放在一个queue里边，然后在一个循环里，每次取出queue中的一个元素i，把它加入结果，更新前置课为i的点 j 的入度（入度-1）。如果点 j 此时的入度0，那么把它加入到队列当中。

为了提高效率，代码里边的result可以替换为一个int count，最后判断count的值是否为nu mCourse ，不是的话说明有环。

代码：

public boolean canFinish(int numCourses, int[][] prerequisites) {        if(numCourses == 0) return true;        int [] inDegree = new int[numCourses];        //init        for(int i=0;i<prerequisites.length;i++){                inDegree[prerequisites[i][0]]++;        }        Queue<Integer> queue = new LinkedList<>();        for(int i=0;i<inDegree.length;i++){            if(inDegree[i] == 0){                queue.offer(i);            }        }        //List<Integer> result = new ArrayList<>();        int count = 0;        while(queue.size() != 0){                        int cur = queue.poll();            //result.add(cur);            count++;            //update            inDegree[cur] = -1;            for(int i=0;i<prerequisites.length;i++){                if(prerequisites[i][1] == cur){                    inDegree[prerequisites[i][0]]--;                    if(inDegree[prerequisites[i][0]]==0){                        queue.offer(prerequisites[i][0]);                    }                }            }        }        if(count == numCourses) return true;        return false;    }