LeetCode Reconstruct Itinerary

原题链接在这里：https://leetcode.com/problems/reconstruct-itinerary/description/

题目：

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.

题解：

Eulerian path. 把这些ticket当成edge构建directed graph. 保证每条edge 只走一遍.

为了保证字母顺序，用了PriorityQueue.

然后做dfs. dfs 时注意 retrieve nodes backwards.

Time Complexity: O(n+e). Space: O(n+e).

AC Java:

 class Solution {
     public List<String> findItinerary(List<List<String>> tickets) {
         List<String> res = new ArrayList<>();
         if(tickets == null || tickets.size() == 0){
             return res;
         }

         HashMap<String, PriorityQueue<String>> graph = new HashMap<>();
         for(List<String> e : tickets){
             graph.putIfAbsent(e.get(0), new PriorityQueue<String>());
             graph.get(e.get(0)).add(e.get(1));
         }

         dfs("JFK", graph, res);
         return res;
     }

     private void dfs(String cur, HashMap<String, PriorityQueue<String>> graph, List<String> res){
         while(graph.containsKey(cur) && graph.get(cur).size() != 0){
             String next = graph.get(cur).poll();
             dfs(next, graph, res);
         }

         res.add(0, cur);
     }
 }