leetcode 332. Reconstruct Itinerary

发表于 | 分类于

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:

  • 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"].
  • All airports are represented by three capital letters (IATA code).
  • You may assume all tickets form at least one valid itinerary.

Example 1:

Input: tickets = [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]
Output: ["JFK", "MUC", "LHR", "SFO", "SJC"]

Example 2:

Input: tickets = [["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.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
import heapq
import collections
class Solution(object):
    def findItinerary(self, tickets):
        """
        :type tickets: List[List[str]]
        :rtype: List[str]
        """
        if not tickets:return []
        dic = collections.defaultdict(list)
        res = []
        for fr,to in tickets:
            heapq.heappush(dic[fr],to)
        def dfs(city):
            if city in dic:
                while dic[city]:
                    dfs(heapq.heappop(dic[city]))
            res.append(city)
        dfs('JFK')
        return res[::-1]