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.

解题思路：

All the airports are vertices and tickets are directed edges. Then all these tickets form a directed graph.

The graph must be Eulerian since we know that a Eulerian path exists.

Thus, start from "JFK", we can apply the Hierholzer's algorithm to find a Eulerian path in the graph which is a valid reconstruction.

Since the problem asks for lexical order smallest solution, we can put the neighbors in a min-heap. In this way, we always visit the smallest possible neighbor first in our trip.

public class Solution {
    public List<String> findItinerary(String[][] tickets) {
        Map<String, PriorityQueue<String>> flights = new HashMap<>();
        LinkedList<String> res = new LinkedList<>();
        for(String[] ticket : tickets) {
            flights.putIfAbsent(ticket[0], new PriorityQueue<String>());
            flights.get(ticket[0]).add(ticket[1]);
        }
        dfs("JFK", flights, res);
        return res;
    }
    
    private void dfs(String departure, Map<String, PriorityQueue<String>> flights, LinkedList<String> res) {
        PriorityQueue<String> arrivals = flights.get(departure);
        while(arrivals != null && !arrivals.isEmpty()) {
            dfs(arrivals.poll(), flights, res);
        }
        res.addFirst(departure);
    }
}