Course Schedule(拓扑）

There are a total ofncourses you have to take, labeled from0ton - 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 prerequisitepairs, 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:

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.

Hints:

1. This problem is equivalent to finding if a cycle exists in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.

2. Topological Sort via DFS

   • A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.

3. Topological sort could also be done via

   BFS

分析

有序图用拓扑排序检查是否有环，无序图用并查集。注意这里课程是反序input。

用一个map存node和它的neighbors们List，一个数组存入度。然后BFS，用一个Queue.最后若是所有数组都遍历到，则无环。

有环的话入度不为0不会被加入queue,所以不可达。

class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        Map<Integer, List<Integer>> neighbors = new HashMap<Integer, List<Integer>>();//node and its neighbors
        Queue<Integer> q = new LinkedList<Integer>();//bfs
        int[] indegree = new int[numCourses];//indegree出度
        int cnt = numCourses;
        for(int[] p : prerequisites){
            indegree[p[0]] ++;
            if(neighbors.containsKey(p[1])){
                neighbors.get(p[1]).add(p[0]);
            }else{
                List<Integer> ns = new ArrayList<Integer>();
                ns.add(p[0]);
                neighbors.put(p[1], ns);
            }
        }
        for(int i = 0; i < indegree.length; i ++){
            if(indegree[i] == 0){
                q.offer(i);
            }
        }
        while(!q.isEmpty()){
            int cur = q.poll();
            cnt --;
            if(neighbors.containsKey(cur)){
              for(int i : neighbors.get(cur)){
                indegree[i] --;
                if(indegree[i] == 0){
                    q.offer(i);
                }
            }
            }

        }
        return cnt == 0;
    }
}