# 210. Course Schedule II

## Description

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, return the ordering of courses you should take to finish all courses.

There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.

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 the correct course order is [0,1]

`4, [[1,0],[2,0],[3,1],[3,2]]`

There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. So one correct course order is [0,1,2,3]. Another correct ordering is[0,2,1,3].

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.

## Answer

This question is a kind of Graph Traversal problem. While meeting two or more nodes become prerequisities of each others in circle, the problem has no answer.

At first, saving the matrix prerequisities to an adjacency list.

Second, enqueue those nodes who have no prerequisite.

By using `indegree`

to record the number of prerequisites of one course, and only all its prerequisites have been taken can enter the course to the queue.

As a result, the problem likes a toposort problem, and I uses Broad-first Search to sovle it.

Code is following:

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public class Solution { public int[] findOrder(int numCourses, int[][] prerequisites) { List<Integer>[] adj = new List[numCourses]; Queue<Integer> queue = new LinkedList<>(); int[] indegree = new int[numCourses]; int[] result = new int[numCourses]; int count = 0; for(int i =0;i<prerequisites.length;i++){ // Using Adjancency list to save edges of a graph List<Integer> list = adj[prerequisites[i][1]]; if(list==null){ list = new LinkedList<>(); adj[prerequisites[i][1]] = list; } adj[prerequisites[i][1]].add(prerequisites[i][0]); indegree[prerequisites[i][0]]++; } for(int i=0;i<numCourses;i++){ // indegree ==0 means no prerequisties if(indegree[i]==0){ queue.offer(i); result[count++]=i; } } while(!queue.isEmpty()){ // a typical BFS int course = queue.poll(); if(adj[course]==null) continue; for(int one:adj[course]){ indegree[one]--; if(indegree[one]==0){ queue.offer(one); result[count++]=one; } } } return (count==numCourses)?result:new int[0]; } } |