• 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 }
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  • 原文地址:https://www.cnblogs.com/lishiblog/p/5744079.html
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