# 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:
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.
```

### Solutions:

``````public class Solution {
public List<String> findItinerary(String[][] tickets) {
HashMap<String, List<String>> flights = new HashMap<String, List<String>>();
for (int i = 0; i < tickets.length; i ++) {
if (!flights.containsKey(tickets[i][0])) {
}
}
for (String c:flights.keySet()) {
Collections.sort(flights.get(c));
}
return fi(tickets.length + 1, "JFK", flights, new LinkedList<String>(), remain);
}
private List<String> fi(int n, String port, HashMap<String, List<String>> flights, List<String> curr, List<String> remain) {
if (curr.size() == n) {
}
if (flights.containsKey(port)) {
for (String next:flights.get(port)) {
if (!remain.contains(port + next)) {
continue;
}
remain.remove(port + next);
List<String> result = fi(n, next, flights, curr, remain);
if (result != null) {
return result;
}
}
}
curr.remove(curr.size() - 1);
return null;
}
}
``````
``````public class Solution {
public List<String> findItinerary(String[][] tickets) {
HashMap<String, PriorityQueue<String>> flights = new HashMap<String, PriorityQueue<String>>();
for (int i = 0; i < tickets.length; i ++) {
if (!flights.containsKey(tickets[i][0])) {
flights.put(tickets[i][0], new PriorityQueue<String>());
}
}
fi(result, "JFK", flights);
return result;
}
private void fi(List<String> result, String port, HashMap<String, PriorityQueue<String>> flights) {
PriorityQueue<String> q = flights.get(port);
while (q != null && !q.isEmpty()) {
fi(result, q.poll(), flights);
}