LeetCode-Reconstruct Itinerary

Given a list of airline tickets represented by pairs of departure and arrival airports [from, to], reconstruct the itinerary in order. All of the tickets belong to a man who departs from JFK. Thus, the itinerary must begin with JFK.

Note:

1. If there are multiple valid itineraries, you should return the itinerary that has the smallest lexical order when read as a single string. For example, the itinerary ["JFK", "LGA"] has a smaller lexical order than ["JFK", "LGB"].
2. All airports are represented by three capital letters (IATA code).
3. You may assume all tickets form at least one valid itinerary.

Example 1:
tickets = [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]
Return ["JFK", "MUC", "LHR", "SFO", "SJC"].

Example 2:
tickets = [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
Return ["JFK","ATL","JFK","SFO","ATL","SFO"].
Another possible reconstruction is ["JFK","SFO","ATL","JFK","ATL","SFO"]. But it is larger in lexical order.

Analysis:

Use DFS. Termination: when all tickets are used. Invalid: if haven't used all ticket, but reach some city without any outgoing tickets.

Solution:

public class Solution {
    public class TicketGroup {
        List<String> outList;
        boolean[] visited;

        public TicketGroup(String firstOutCity){
            outList = new ArrayList<String>();
            outList.add(firstOutCity);
        }        
    }
    
    public List<String> findItinerary(String[][] tickets) {
        List<String> resList = new ArrayList<String>();
        if (tickets.length == 0) return resList;

        HashMap<String,TicketGroup> graph = new HashMap<String,TicketGroup>();
        for (String[] ticket : tickets){
            if (graph.containsKey(ticket[0])){
                graph.get(ticket[0]).outList.add(ticket[1]);
            } else {
                TicketGroup group = new TicketGroup(ticket[1]);
                graph.put(ticket[0],group);
            }
        }
        for (TicketGroup group : graph.values()){
            Collections.sort(group.outList);
            group.visited = new boolean[group.outList.size()];
        }

        resList.add("JFK");
        findItineraryRecur("JFK",graph,resList,tickets.length+1);
        return resList;
    }

    public boolean findItineraryRecur(String cur, HashMap<String,TicketGroup> graph, List<String> resList, int maxLen){
        if (resList.size() == maxLen)
            return true;

        if (!graph.containsKey(cur)) return false;
        boolean available = false;
        TicketGroup curGroup = graph.get(cur);
        for (boolean visit : curGroup.visited)
            if (!visit){
                available = true;
                break;
            }
        if (!available) return false;

        for (int i=0;i<curGroup.visited.length;i++){
            if (!curGroup.visited[i]){
                curGroup.visited[i] = true;
                resList.add(curGroup.outList.get(i));
                if (findItineraryRecur(curGroup.outList.get(i),graph,resList,maxLen)){
                    return true;
                }
                resList.remove(resList.size()-1);
                curGroup.visited[i] = false;
            }
        }
        return false;    
    }
}