332. 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.

一刷
思路， 建立一个map, key为departure airport, value为 arrival airports的priority queue。可以保证dequeue时的 lexical order。然后做dfs

public class Solution {
    Map<String, PriorityQueue<String>> targets = new HashMap<>();
    List<String> route = new LinkedList();
    
    public List<String> findItinerary(String[][] tickets) {
        for(String[] ticket : tickets){
            targets.computeIfAbsent(ticket[0], k->new PriorityQueue()).add(ticket[1]);
        }
        visit("JFK");
        return route;
    }
    
    private void visit(String airport){
        while(targets.containsKey(airport) && !targets.get(airport).isEmpty())
            visit(targets.get(airport).poll());
        route.add(0, airport);
    }
}