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.
一刷
思路， 建立一个map, key为departure airport, value为 arrival airports的priority queue。可以保证dequeue时的 lexical order。然后做dfs
public class Solution {
Map<String, PriorityQueue<String>> targets = new HashMap<>();
List<String> route = new LinkedList();
public List<String> findItinerary(String[][] tickets) {
for(String[] ticket : tickets){
targets.computeIfAbsent(ticket[0], k->new PriorityQueue()).add(ticket[1]);
}
visit("JFK");
return route;
}
private void visit(String airport){
while(targets.containsKey(airport) && !targets.get(airport).isEmpty())
visit(targets.get(airport).poll());
route.add(0, airport);
}
}
