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.

Solution:

public class Solution {
    public List<String> findItinerary(String[][] tickets) {
        Map<String,PriorityQueue<String>> map = new HashMap<>();
        for(String[] ti : tickets){
            if(!map.containsKey(ti[0])) map.put(ti[0], new PriorityQueue<>());
            map.get(ti[0]).offer(ti[1]);
        }
        Stack<String> stack = new Stack<>();
        List<String> ans = new ArrayList<>();
        stack.push("JFK");
        while(!stack.isEmpty()){
            String cur = stack.peek();
            if(map.containsKey(cur) && !map.get(cur).isEmpty()){
                stack.push(map.get(cur).poll());
            } else {
                stack.pop();
                ans.add(0, cur);
            }
        }
        return ans;
    }
}