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 departs from JFK . Thus, the itinerary must begin with JFK .

Note:

1. If There is multiple valid itineraries, you should return the itinerary that have the smallest lexical order when read as A single string. For example, the itinerary have ["JFK", "LGA"] a smaller lexical order than ["JFK", "LGB"] .
2. All airports is represented by three capital letters (IATA code).
3. 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's larger in lexical order.

Analysis:

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

Solution:

1  Public classSolution {2      Public classTicketgroup {3List<string>outlist;4         Boolean[] visited;5 6          PublicTicketgroup (String firstoutcity) {7Outlist =NewArraylist<string>();8 Outlist.add (firstoutcity);9         }        Ten     } One      A      PublicList<string>finditinerary (string[][] tickets) { -List<string> reslist =NewArraylist<string>(); -         if(Tickets.length = = 0)returnreslist; the  -hashmap<string,ticketgroup> graph =NewHashmap<string,ticketgroup>(); -          for(string[] ticket:tickets) { -             if(Graph.containskey (ticket[0])){ +Graph.get (Ticket[0]). Outlist.add (ticket[1]); -}Else { +Ticketgroup Group =NewTicketgroup (ticket[1]); AGraph.put (ticket[0],group); at             } -         } -          for(Ticketgroup group:graph.values ()) { - Collections.sort (group.outlist); -group.visited =New Boolean[Group.outList.size ()]; -         } in  -Reslist.add ("JFK"); toFinditineraryrecur ("JFK", graph,reslist,tickets.length+1); +         returnreslist; -     } the  *      Public BooleanFinditineraryrecur (String cur, hashmap<string,ticketgroup> graph, list<string> reslist,intmaxlen) { $         if(reslist.size () = =maxlen)Panax Notoginseng             return true; -  the         if(!graph.containskey (cur))return false; +         BooleanAvailable =false; ATicketgroup Curgroup =graph.get (cur); the          for(Booleanvisit:curGroup.visited) +             if(!visit) { -Available =true; $                  Break; $             } -         if(!available)return false; -  the          for(inti=0;i<curgroup.visited.length;i++){ -             if(!Curgroup.visited[i]) {WuyiCurgroup.visited[i] =true; the Reslist.add (CurGroup.outList.get (i)); -                 if(Finditineraryrecur (CurGroup.outList.get (i), Graph,reslist,maxlen)) { Wu                     return true; -                 } AboutReslist.remove (Reslist.size ()-1); $Curgroup.visited[i] =false; -             } -         } -         return false;  A     } +}

Leetcode-reconstruct Itinerary