leetcode [332]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 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: [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]
Output: ["JFK", "MUC", "LHR", "SFO", "SJC"]

Example 2:

Input: [["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.

题目大意：

给一连串起始地和目的地，输出最后的路线。

解法：

采用深度优先遍历。

java:

class Solution {
    Map<String,PriorityQueue<String>>targets=new HashMap<>();
    List<String>route=new LinkedList<>();

    private void visit(String airport){
        while (targets.containsKey(airport)&&!targets.get(airport).isEmpty())
            visit(targets.get(airport).poll());
        route.add(0,airport);
    }
    
    public List<String> findItinerary(List<List<String>> tickets) {
        for (final List<String>ticket:tickets)
            targets.computeIfAbsent(ticket.get(0),k->new PriorityQueue<String>()).add(ticket.get(1));
        visit("JFK");
        
        return route;
    }
}