def test_append_when_list_is_empty(self):
singlyLinkedListObject = SinglyLinkedList()
inputVal = 10
singlyLinkedListObject.appendToListUsingTail(inputVal)
outputList = singlyLinkedListObject.returnLinkedListAsList()
outputVal = outputList[0]
self.assertTrue(inputVal == outputVal, "Expected a tail value of: %s, Got: %s" %(inputVal, outputVal))
def test_append_when_list_has_at_least_one_element(self):
singlyLinkedListObject = SinglyLinkedList()
inputList = range(10)
singlyLinkedListObject.populate(inputList)
inputVal = 10
singlyLinkedListObject.appendToListUsingTail(inputVal)
self.assertTrue(singlyLinkedListObject.tail.data == inputVal, "Expected Tail data: %s, Got: %s" %(inputVal, singlyLinkedListObject.tail.data))
outputList = singlyLinkedListObject.returnLinkedListAsList()
outputVal = outputList[-1]
self.assertTrue(inputVal == outputVal, "Expected tail value: %s, Got: %s" %(inputVal, outputVal))
def test_changing_list_length_during_append_operation(self):
singlyLinkedListObject = SinglyLinkedList()
expectedLength = 0
actualLength = len(singlyLinkedListObject)
self.assertTrue(expectedLength == actualLength, "Expected length: %s, Actual Length: %s" %(expectedLength, actualLength))
inputList = range(10)
for data in inputList:
singlyLinkedListObject.appendToListUsingTail(data)
actualLength = len(singlyLinkedListObject)
expectedLength += 1 # Because a new element has been added to the list.
self.assertTrue(expectedLength == actualLength, "Expected length: %s, Actual Length: %s" %(expectedLength, actualLength))
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
def addTwoReversedIntegersAsLinkedListsAndCreateNewList(self):
outputList = SinglyLinkedList()
carry = 0
firstListCurrentNode = self.firstList.head
secondListCurrentNode = self.secondList.head
while firstListCurrentNode is not None and secondListCurrentNode is not None:
sum = carry + firstListCurrentNode.data + secondListCurrentNode.data
data = sum % 10
carry = sum / 10
# print "data: %s" %data
# print "carry: %s" %carry
outputList.appendToListUsingTail(data)
firstListCurrentNode = firstListCurrentNode.nextPointer
secondListCurrentNode = secondListCurrentNode.nextPointer
# there could be a left-over carry. Need to add it to the list now, but before that, need to check if the number of items in the list were uneven
currentNode = None
if firstListCurrentNode is None:
currentNode = secondListCurrentNode
else:
currentNode = firstListCurrentNode
# If both firstListCurrentNode and secondListCurrentNode are None, it is implicitly covered
while currentNode is not None:
sum = carry + currentNode.data
data = sum % 10
carry = sum / 10
outputList.appendToListUsingTail(data)
currentNode = currentNode.nextPointer
# Now, check the left-over carry and add it to the list if it is non-zero
if carry != 0:
outputList.appendToListUsingTail(carry)
# print outputList.returnLinkedListAsList()
return outputList