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 fromJFK. 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.

思路：
这个题是一个图的题目

1. 首先建立图，利用hashmap和priorityqueue，因为题目要求字母最小序列，那么用pq，可以利用string自己本身的compareTo功能，进行排序.
2. 从JFK开始dfs，每次找到终点，那么都要从PriorityQueue里面弹出来(每次访问过的就会从hashmap中弹出来），然后反向加回来（每一次都加在list的最前面），在dfs的最后加过来就是正的顺序。

class Solution {
    public List findItinerary(String[][] tickets) {
        List path = new ArrayList();
        
        if (tickets == null || tickets.length == 0) {
            return path;
        }
        
        //1. reconstruct the graph based on priorityQueue
        HashMap> mapping = new HashMap<>();
        
        for (int i = 0; i < tickets.length; i++) {
            if (!mapping.containsKey(tickets[i][0])) {
                //PriorityQueue queue = new PriorityQueue(); 
                mapping.put(tickets[i][0], new PriorityQueue());
            }
            mapping.get(tickets[i][0]).offer(tickets[i][1]);
        }
        
        //2. DFS to find all path
        findPath(mapping, path, "JFK");
        return path;
    } 
    
    public void findPath(HashMap> mapping, List path, String startWord) {
        PriorityQueue queue = mapping.get(startWord);
        while (queue != null && !queue.isEmpty()) {
            findPath(mapping, path, queue.poll());
        }
        path.add(0, startWord);
    }
}