You are given a list of airline tickets
where tickets[i] = [fromi, toi]
represent the departure and the arrival airports of one flight. Reconstruct the itinerary in order and return it.
All of the tickets belong to a man who departs from "JFK"
, thus, the itinerary must begin with "JFK"
. 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"]
.
You may assume all tickets form at least one valid itinerary. You must use all the tickets once and only once.
Input: tickets = [["MUC","LHR"],["JFK","MUC"],["SFO","SJC"],["LHR","SFO"]] Output: ["JFK","MUC","LHR","SFO","SJC"]
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 <= tickets.length <= 300
tickets[i].length == 2
fromi.length == 3
toi.length == 3
fromi
andtoi
consist of uppercase English letters.fromi != toi
classSolution: deffindItinerary(self, tickets: List[List[str]]) ->List[str]: targets= {} stack= [] used=set() itinerary= [("JFK", None, None)] forfr0m, tointickets: iffr0mnotintargets: targets[fr0m] = [] iftonotintargets: targets[to] = [] targets[fr0m].append(to) forfr0mintargets: targets[fr0m].sort(reverse=True) whileTrue: fr0m=itinerary[-1][0] foriinrange(len(targets[fr0m])): if (fr0m, i) notinused: ifstack== [] \ orfr0m!=stack[-1][0] \ ortargets[fr0m][i] !=targets[fr0m][stack[-1][1]] \ orlen(itinerary) !=stack[-1][2]: stack.append((fr0m, i, len(itinerary))) whilestack[-1][2] <len(itinerary): _, fr0m, i=itinerary.pop() used.remove((fr0m, i)) fr0m, i, _=stack.pop() used.add((fr0m, i)) itinerary.append((targets[fr0m][i], fr0m, i)) iflen(itinerary) ==len(tickets) +1: return [airportforairport, _, _initinerary]