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 fromJFK. Thus, the itinerary must begin withJFK.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:
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 is larger in lexical order.The Idea: This is a eularian path/cycle problem. Our goal is to visit every edge of the graph once, and return to our original position (in our case, which is '
JFK'). The algorithm that I have used to implement this idea is called Hierholzer's Algorithm. More details and discussion here: https://maksimdan.gitbooks.io/ecs122a-algorithm-design-lecture-notes/content/eularian-cycle.htmlComplexity: O(|E|) time and O(|E|) space
from collections import defaultdict
class Solution:
def findItinerary(self, tickets):
"""
:type tickets: List[List[str]]
:rtype: List[str]
"""
if not tickets or not tickets[0]:
return []
# build representative graph
g = defaultdict(list)
for _to, _from in tickets:
g[_to].append(_from)
# now sort to create smallest
# lexicographical itinerary
for _to, _ in g.items():
g[_to].sort()
# maintain an index where the eularian path
# left off
left_off = {}
for _to, _from in tickets:
left_off[_to] = 0
left_off[_from] = 0
itinerary = []
s = ['JFK']
# now find the eularian path
while s:
top = s[-1]
g_index = left_off[top]
g_list = g[top]
if g_index < len(g_list):
s.append(g_list[g_index])
left_off[top] = g_index + 1
else:
itinerary.append(top)
s.pop()
return list(reversed(itinerary))
Last modified 4yr ago