def partitionLinkedListBasedOnAGivenNumber_createNewList(self, givenNumber):
firstList = SinglyLinkedList()
secondList = SinglyLinkedList()
for data in self.givenSinglyLinkedList:
if data < givenNumber:
firstList.appendToListUsingTail(data)
else:
secondList.appendToListUsingTail(data)
# print firstList.returnLinkedListAsList()
# print secondList.returnLinkedListAsList()
if firstList.isListEmpty():
return secondList
elif secondList.isListEmpty():
return firstList
else:
firstList.tail.nextPointer = secondList.head # Could be the other way round as well (secondList.tail.nextPointer = firstList.head ), but since it wasn't asked in the question, I've chosen the first option. In a real interview, I would ask the interviewer
return firstList
def partitionLinkedListBasedOnAGivenNumber_createNewList(
self, givenNumber):
firstList = SinglyLinkedList()
secondList = SinglyLinkedList()
for data in self.givenSinglyLinkedList:
if data < givenNumber:
firstList.appendToListUsingTail(data)
else:
secondList.appendToListUsingTail(data)
# print firstList.returnLinkedListAsList()
# print secondList.returnLinkedListAsList()
if firstList.isListEmpty():
return secondList
elif secondList.isListEmpty():
return firstList
else:
firstList.tail.nextPointer = secondList.head # Could be the other way round as well (secondList.tail.nextPointer = firstList.head ), but since it wasn't asked in the question, I've chosen the first option. In a real interview, I would ask the interviewer
return firstList
def partitionLinkedListBasedOnAGivenNumber_DontCreateNewList(self, givenNumber):
# check head against given number. Move to appropriate list. advance head to next. delete previous head
firstList = SinglyLinkedList()
secondList = SinglyLinkedList()
for data in self.givenSinglyLinkedList:
if data < givenNumber:
firstList.appendToListUsingTail(data)
else:
secondList.appendToListUsingTail(data)
self.givenSinglyLinkedList.deleteHead() # Deleting head as the data is already under the relevant - firstList or secondList
if firstList.isListEmpty():
return secondList
elif secondList.isListEmpty():
return firstList
else:
firstList.tail.nextPointer = secondList.head # Could be the other way round as well (secondList.tail.nextPointer = firstList.head ), but since it wasn't asked in the question, I've chosen the first option. In a real interview, I would ask the interviewer
return firstList
def partitionLinkedListBasedOnAGivenNumber_DontCreateNewList(
self, givenNumber):
# check head against given number. Move to appropriate list. advance head to next. delete previous head
firstList = SinglyLinkedList()
secondList = SinglyLinkedList()
for data in self.givenSinglyLinkedList:
if data < givenNumber:
firstList.appendToListUsingTail(data)
else:
secondList.appendToListUsingTail(data)
self.givenSinglyLinkedList.deleteHead(
) # Deleting head as the data is already under the relevant - firstList or secondList
if firstList.isListEmpty():
return secondList
elif secondList.isListEmpty():
return firstList
else:
firstList.tail.nextPointer = secondList.head # Could be the other way round as well (secondList.tail.nextPointer = firstList.head ), but since it wasn't asked in the question, I've chosen the first option. In a real interview, I would ask the interviewer
return firstList