Leetcode Reconstruct itinerary

The original title link is here: https://leetcode.com/problems/reconstruct-itinerary/

Topic:

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 departs from JFK . Thus, the itinerary must begin with JFK .

Note:

1. If There is multiple valid itineraries, you should return the itinerary that have the smallest lexical order when read as A single string. For example, the itinerary have ["JFK", "LGA"] a smaller lexical order than ["JFK", "LGB"] .
2. All airports is represented by three capital letters (IATA code).
3. 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's larger in lexical order.

Exercises

Build directed graph with these ticket as edge. In order to guarantee the alphabetical order, Priorityqueue was used. Then do DFS.

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

AC Java:

1  Public classSolution {2map<string, priorityqueue<string>> graph =NewHashmap<string, priorityqueue<string>>();3      PublicList<string>finditinerary (string[][] tickets) {4list<string> res =NewLinkedlist<string>();5         if(Tickets = =NULL|| Tickets.length = = 0 | | Tickets[0].length = = 0){6             returnRes;7         }8         9          for(String [] edge:tickets) {Ten             if(!graph.containskey (edge[0])){ OneGraph.put (Edge[0],NewPriorityqueue<string>()); A             } -Graph.get (Edge[0]). Add (edge[1]); -         } the          -DFS ("JFK", res); -         returnRes; -     } +      -     Private voidDFS (String s, list<string>Res) { +          while(Graph.containskey (s) &&!Graph.get (s). IsEmpty ()) { A Dfs (Graph.get (s). Poll (), res); at         } -Res.add (0, s); -     } -}

Leetcode Reconstruct itinerary