LeetCode: Reconstruct Itinerary

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.

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public class Solution {
          private LinkedList<String> result;
    private Map<String, PriorityQueue<String>> map;

    public List<String> findItinerary(String[][] tickets) {
        if (tickets == null || tickets.length == 0) {
            return Collections.emptyList();
        }

        result = new LinkedList<>();
        map = new HashMap<>();
        for (String[] ticket : tickets) {
            String from = ticket[0];
            if (!map.containsKey(from)) {
                map.put(from, new PriorityQueue<>());
            }
            map.get(from).add(ticket[1]);
        }
        dfs("JFK");
        return result;
    }

    private void dfs(String start) {
        while (map.containsKey(start) && !map.get(start).isEmpty()) {
            dfs(map.get(start).poll());
        }
        result.addFirst(start);
    }
}