[LeetCode] 332. Reconstruct Itinerary

Problem

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:

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.

Solution

class 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);
    }
}