-
Notifications
You must be signed in to change notification settings - Fork 1
/
Copy path332 Reconstruct Itinerary.py
56 lines (43 loc) · 2.02 KB
/
332 Reconstruct Itinerary.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
"""
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:
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.
Author: Rajeev Ranjan
"""
import heapq
from collections import defaultdict, deque
class Solution(object):
def findItinerary(self, tickets):
"""
Euler path:
An Euler path is a path that uses every edge of a graph exactly once.
Hierholzer's algorithm a Euler path, must be directed graph
The graph must be directed graph
Heap can be replaced by stack/queue and sort the original list
The ret is build as from right to left: JFK <- NRT <- JFK <- KUL
:type tickets: List[List[str]]
:rtype: List[str]
"""
G = defaultdict(list) # every list is a heap
for s, e in tickets:
heapq.heappush(G[s], e) # heap lexical order
ret = deque()
self.dfs(G, 'JFK', ret)
return list(ret)
def dfs(self, G, cur, ret):
while G[cur]:
self.dfs(G, heapq.heappop(G[cur]), ret)
ret.appendleft(cur)
if __name__ == "__main__":
assert Solution().findItinerary([["JFK","KUL"],["JFK","NRT"],["NRT","JFK"]]) == ['JFK', 'NRT', 'JFK', 'KUL']