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:

Input: tickets = [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]
Output: ["JFK", "MUC", "LHR", "SFO", "SJC"]

Example 2:

Input: tickets = [["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.

学习来offer的代码。
典型图和dfs问题。注意数据类型PriorityQueue来解决Lexical order排序，很省事。

class Solution {
    public List<String> findItinerary(String[][] tickets) {
        Map<String, PriorityQueue<String>> map = new HashMap<>();
        LinkedList<String> result = new LinkedList<>();
        for(String[] ticket: tickets){
            if(!map.containsKey(ticket[0])){
                PriorityQueue<String> q = new PriorityQueue<>();
                map.put(ticket[0],q);
            }
            map.get(ticket[0]).offer(ticket[1]);
        }
        dfs("JFK",result,map);
        return result;
    }
    private void dfs(String s, LinkedList<String> result, Map<String, PriorityQueue<String>> map){
        PriorityQueue<String> q = map.get(s);
        while(q!=null && !q.isEmpty()){
            dfs(q.poll(),result,map);
        }
        result.addFirst(s);
    }
}