題目：

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.

題意及分析：將所有的機票用數組保存起來，最後我們就是要找出一條路徑將機票用完，並且如果有多條路徑，就找字典序最小的。最開始想到的方法是用一個hashmap<String,PriorityQueue<String>>保存每個點及其鄰節點，然後使用深度遍歷，第一次得到的結果便是答案（因為每次都是用的最小路徑）。 另外一種方法是每次找到終點，然後刪除該點繼續查找下一個終點，最後得到的結果反轉即可。

代碼：

public class Solution {
    Map<String, PriorityQueue<String>>  map= new HashMap<>();
    List<String> res = new ArrayList<>();

    public List<String> findItinerary(String[][] tickets) {
        for(String[] ticket:tickets){
            map.computeIfAbsent(ticket[0],k->new PriorityQueue<>()).add(ticket[1]);
        }
        dfs("JFK");
        Collections.reverse(res);
        return res;
    }

    public void dfs(String node){
        PriorityQueue<String> priorityQueue = map.get(node);
        while(priorityQueue!=null&&!priorityQueue.isEmpty())
            dfs(priorityQueue.poll());
        res.add(node);
    }
}

[LeetCode] 332. Reconstruct Itinerary Java