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.

click to show more hints.

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. There are several ways to represent a graph. For example, the input prerequisites is a graph represented by a list of edges. Is this graph representation appropriate?
3. Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
4. Topological sort could also be done via BFS.

 

这题的重点在于找到不depend on any node的node， 然后一层一层的降depend。 

时间复杂度是O(n^2) n为node的个数

空间复杂度是O(e) e 为depend 的个数（即： 边数）

 1 public class Solution {

 2     public boolean canFinish(int numCourses, int[][] prerequisites) {

 3         if(prerequisites == null || prerequisites.length == 0 || prerequisites[0].length != 2){

 4             return true;

 5         }

 6         int[] depend = new int[numCourses];//node[i] depends on depend[i] # of nodes

 7         //arraylist of node depends on node key

 8         HashMap<Integer, ArrayList<Integer>> mapping = new HashMap<Integer, ArrayList<Integer>>();

 9         for(int i = 0; i < prerequisites.length; i ++){

10             depend[prerequisites[i][0]] ++;

11             if(mapping.containsKey(prerequisites[i][1])){

12                 mapping.get(prerequisites[i][1]).add(prerequisites[i][0]);

13             }else{

14                 ArrayList<Integer> list = new ArrayList<Integer>();

15                 list.add(prerequisites[i][0]);

16                 mapping.put(prerequisites[i][1], list);

17             }

18         }

19         while(mapping.size() != 0){

20             boolean flg = false;

21             for(int i = 0; i < numCourses; i ++){

22                 if(depend[i] == 0 && mapping.containsKey(i)){

23                     //this node doesn't depend on any node, but some node depend on it

24                     flg = true;

25                     for(Integer pos : mapping.get(i)){

26                         depend[pos] --;

27                     }

28                     mapping.remove(i);

29                 }

30             }

31             if(flg == false){

32                 return false;

33             }

34         }

35         return true;

36     }

37 }