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）

public class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        if(prerequisites == null || prerequisites.length == 0 || prerequisites[0].length != 2){
            return true;
        }
        int[] depend = new int[numCourses];//node[i] depends on depend[i] # of nodes
        //arraylist of node depends on node key
        HashMap<Integer, ArrayList<Integer>> mapping = new HashMap<Integer, ArrayList<Integer>>();
        for(int i = 0; i < prerequisites.length; i ++){
            depend[prerequisites[i][0]] ++;
            if(mapping.containsKey(prerequisites[i][1])){
                mapping.get(prerequisites[i][1]).add(prerequisites[i][0]);
            }else{
                ArrayList<Integer> list = new ArrayList<Integer>();
                list.add(prerequisites[i][0]);
                mapping.put(prerequisites[i][1], list);
            }
        }
        while(mapping.size() != 0){
            boolean flg = false;
            for(int i = 0; i < numCourses; i ++){
                if(depend[i] == 0 && mapping.containsKey(i)){
                    //this node doesn't depend on any node, but some node depend on it
                    flg = true;
                    for(Integer pos : mapping.get(i)){
                        depend[pos] --;
                    }
                    mapping.remove(i);
                }
            }
            if(flg == false){
                return false;
            }
        }
        return true;
    }
}