[LeetCode] 332. Reconstruct Itinerary

ProblemGiven 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: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”].All airports are represented by three capital letters (IATA code).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.Solutionclass Solution { public List<String> findItinerary(String[][] tickets) { Map<String, PriorityQueue<String>> map = new HashMap<>(); List<String> res = new ArrayList<>(); for (String[] ticket: tickets) { if (!map.containsKey(ticket[0])) map.put(ticket[0], new PriorityQueue<>()); map.get(ticket[0]).add(ticket[1]); } dfs(“JFK”, map, res); return res; } public void dfs(String departure, Map<String, PriorityQueue<String>> map, List<String> res) { PriorityQueue<String> arrivals = map.get(departure); while (arrivals != null && arrivals.size() > 0) { dfs(arrivals.poll(), map, res); } res.add(0, departure); }}