[LC] 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"]


class Solution {
    public List<String> findItinerary(List<List<String>> tickets) {
        Map<String, PriorityQueue<String>> map = new HashMap<>();
        for (int i = 0; i < tickets.size(); i++) {
            List<String> curList = tickets.get(i);
            if (!map.containsKey(curList.get(0))) {
                map.put(curList.get(0), new PriorityQueue<>());
            }
            map.get(curList.get(0)).add(curList.get(1));
        }
        List<String> res = new LinkedList<>();
        find("JFK", map, res);
        return res;
    }
    
    private void find(String start, Map<String, PriorityQueue<String>> map, List<String> res) {
        while (map.containsKey(start) && !map.get(start).isEmpty()) {
            String cur = map.get(start).poll();
            find(cur, map, res);
        }
        res.add(0, start);
    }
}

 

posted @ 2020-06-07 05:44  xuan_abc  阅读(89)  评论(0编辑  收藏  举报