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.

Have you met this question in a real interview? Yes
Example
For graph as follow:

picture

The topological order can be:

[0, 1, 2, 3, 4, 5]
[0, 2, 3, 1, 5, 4]
…

/** * Definition for Directed graph. * class DirectedGraphNode { *     int label; *     ArrayList<DirectedGraphNode> neighbors; *     DirectedGraphNode(int x) { label = x; neighbors = new ArrayList<DirectedGraphNode>(); } * }; */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();//hash  表        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()) {     //队列            DirectedGraphNode node = q.poll();            for (DirectedGraphNode n : node.neighbors) {                map.put(n, map.get(n) - 1);                if (map.get(n) == 0) {//入度为零时  弹出                    result.add(n);                    q.offer(n);                }            }        }        return result;    }}