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.
- One must use all the tickets once and only once.
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:
We will use the Eulerian Path algorithm.
Step 1: Build Graph using HashMap and store destinations using PriorityQueue, to maintain the lexical order of destinations.
Step 2: Perform a depth-first search and add the final destination first to the result and return the result.
class HackerHeap {
HashMap<String, PriorityQueue<String>> map = new HashMap<String,PriorityQueue<String>>();
LinkedList<String> result = new LinkedList<String>();
public List<String> findItinerary(List<List<String>> tickets) {
// Building The Graph
for(List<String> ticket: tickets){
if(!map.containsKey(ticket.get(0))) {
PriorityQueue<String> q = new PriorityQueue<String>();
map.put(ticket.get(0),q);
}
map.get(ticket.get(0)).offer(ticket.get(1));
}
dfs("JFK");
return result;
}
public void dfs(String s) {
PriorityQueue<String> q = map.get(s);
while(q!=null && !q.isEmpty()) {
dfs(q.poll());
}
result.addFirst(s);
}
}