Reconstruct Itinerary Leetcode


#1

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:

Input: [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]
Output: ["JFK", "MUC", "LHR", "SFO", "SJC"]

Example 2:

Input: [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
Output: ["JFK","ATL","JFK","SFO","ATL","SFO"]
Explanation: Another possible reconstruction is ["JFK","SFO","ATL","JFK","ATL","SFO"].
             But it is larger in lexical order.

#2

The below code uses Hierholzer’s Algorithm to visit all nodes(destinations) exactly once.

class Solution {
    
    LinkedList<String> result=new LinkedList<>();
    HashMap<String,PriorityQueue<String>> map=new HashMap<>();
    
    public List<String> findItinerary(String[][] tickets) {
        for(String[] ticket:tickets)
        {
            if(!map.containsKey(ticket[0]))
            {
                map.put(ticket[0],new PriorityQueue<String>());
            }
            map.get(ticket[0]).offer(ticket[1]);
        }
        helper("JFK");
        return result;
    }
    
    private void helper(String ticket){
        PriorityQueue<String> arrivals = map.get(ticket);
        while(arrivals!=null && arrivals.size()>0){
            helper(arrivals.poll());
        }
        result.addFirst(ticket);
    }
}