Topological Sorting

Source

• lintcode: Topological Sorting

Given an directed graph, a topological order of the graph nodes is defined as follow:

For each directed edge A -> B in graph, A must before B in the order list.
The first node in the order can be any node in the graph with no nodes direct to it.
Find any topological order for the given graph.

Example For graph as follow:

The topological order can be:
[0, 1, 2, 3, 4, 5]
[0, 2, 3, 1, 5, 4]
...
Note
You can assume that there is at least one topological order in the graph.

Challenge
Can you do it in both BFS and DFS?

Java

public class Solution {
    /**
     * @param graph: A list of Directed graph node
     * @return: Any topological order for the given graph.
     */    
    public ArrayList<DirectedGraphNode> topSort(ArrayList<DirectedGraphNode> graph) {
        // write your code here
        ArrayList<DirectedGraphNode> result = new ArrayList<DirectedGraphNode>();
        HashMap<DirectedGraphNode, Integer> map = new HashMap();
        for (DirectedGraphNode node : graph) {
            for (DirectedGraphNode neighbor : node.neighbors) {
                if (map.containsKey(neighbor)) {
                    map.put(neighbor, map.get(neighbor) + 1);
                } else {
                    map.put(neighbor, 1); 
                }
            }
        }
        Queue<DirectedGraphNode> q = new LinkedList<DirectedGraphNode>();
        for (DirectedGraphNode node : graph) {
            if (!map.containsKey(node)) {
                q.offer(node);
                result.add(node);
            }
        }
        while (!q.isEmpty()) {
            DirectedGraphNode node = q.poll();
            for (DirectedGraphNode n : node.neighbors) {
                map.put(n, map.get(n) - 1);
                if (map.get(n) == 0) {
                    result.add(n);
                    q.offer(n);
                }
            }
        }
        return result;
    }
}