Lintcode - Topological Sorting

Thoughts:

1. For each node in the graph, construct a map with node as key, and number of parent nodes as value

2. Looping through left nodes and see if its indegree is 0: if so, remove the node from graph and add it to result; also its neighbors indegree--

A problem while implementing #2 is ConcurrentModificatoinException that I tried to remove the node from map while looping through it. A work around is looping through remaining nodes from graph and remove it from graph directly. Entries in map are never removed.

    public ArrayList<DirectedGraphNode> topSort(ArrayList<DirectedGraphNode> graph) {        ArrayList<DirectedGraphNode> result = new ArrayList<DirectedGraphNode>();        if (graph == null) {            return result;        }        Map<DirectedGraphNode, Integer> map = new HashMap<DirectedGraphNode, Integer>();        for (DirectedGraphNode node : graph) {            map.put(node, 0);        }                for (DirectedGraphNode node : graph) {            for (DirectedGraphNode neighbor : node.neighbors) {                map.put(neighbor, map.get(neighbor)+1);            }        }                while (!graph.isEmpty()) {            int index = 0;            while (index < graph.size()) {                DirectedGraphNode node = graph.get(index);                if (map.get(node).equals(0)) {                    result.add(node);                    graph.remove(node);                    for (DirectedGraphNode neighbor : node.neighbors) {                        map.put(neighbor, map.get(neighbor)-1);                    }                } else {                    index++;                }            }        }        return result;    }