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:

 1 public class Solution {
 2     public class TicketGroup {
 3         List<String> outList;
 4         boolean[] visited;
 5 
 6         public TicketGroup(String firstOutCity){
 7             outList = new ArrayList<String>();
 8             outList.add(firstOutCity);
 9         }        
10     }
11     
12     public List<String> findItinerary(String[][] tickets) {
13         List<String> resList = new ArrayList<String>();
14         if (tickets.length == 0) return resList;
15 
16         HashMap<String,TicketGroup> graph = new HashMap<String,TicketGroup>();
17         for (String[] ticket : tickets){
18             if (graph.containsKey(ticket[0])){
19                 graph.get(ticket[0]).outList.add(ticket[1]);
20             } else {
21                 TicketGroup group = new TicketGroup(ticket[1]);
22                 graph.put(ticket[0],group);
23             }
24         }
25         for (TicketGroup group : graph.values()){
26             Collections.sort(group.outList);
27             group.visited = new boolean[group.outList.size()];
28         }
29 
30         resList.add("JFK");
31         findItineraryRecur("JFK",graph,resList,tickets.length+1);
32         return resList;
33     }
34 
35     public boolean findItineraryRecur(String cur, HashMap<String,TicketGroup> graph, List<String> resList, int maxLen){
36         if (resList.size() == maxLen)
37             return true;
38 
39         if (!graph.containsKey(cur)) return false;
40         boolean available = false;
41         TicketGroup curGroup = graph.get(cur);
42         for (boolean visit : curGroup.visited)
43             if (!visit){
44                 available = true;
45                 break;
46             }
47         if (!available) return false;
48 
49         for (int i=0;i<curGroup.visited.length;i++){
50             if (!curGroup.visited[i]){
51                 curGroup.visited[i] = true;
52                 resList.add(curGroup.outList.get(i));
53                 if (findItineraryRecur(curGroup.outList.get(i),graph,resList,maxLen)){
54                     return true;
55                 }
56                 resList.remove(resList.size()-1);
57                 curGroup.visited[i] = false;
58             }
59         }
60         return false;    
61     }
62 }

 

posted @ 2016-08-06 14:41  LiBlog  阅读(191)  评论(0)  编辑  收藏  举报