#extra while loop condition because it's possible that in the process of deletion we changed the next node to be none and
# it naively changes it to none
while currentNode is not None and currentNode.next is not None:
searchNode = currentNode
while searchNode.next is not None:
if searchNode.next.data == currentNode.data:
searchNode.next = searchNode.next.next
else:
searchNode = searchNode.next
currentNode = currentNode.next
print("===========Test 1==========")
head = Node(4)
head.createLinkedListFromList([5, 6, 4])
head.printLinkedList()
removeDup(head)
print("Remove Duplicates")
head.printLinkedList()
print("===========Test 2==========")
head2 = Node(1)
head2.createLinkedListFromList([1, 1, 1, 1, 2, 2, 3, 3])
head2.printLinkedList()
print("Remove Duplicates")
removeDup(head2)
head2.printLinkedList()
print("===========Test 3==========")
head3 = Node(4)
head3.createLinkedListFromList([5, 6, 4])
#much better version
#remember that deleting a node is just rearranging its next pointer to skip the one we want to delete
#also I don't have to shift the whole linked list
#finally and most importantly, look at the linked list as a whole. Realize that what we were trying to achieve didn't require shifting the whole array
# ex. 1->2->3->4->5 and we want to delete 3 then we would get 1->2->4->5
# from this example we should realize that the 4 is in the old place of the three!
def better_del_middle_node(mNode):
if mNode is None or mNode.data == None:
return False
mNode.data = mNode.next.data
mNode.next = mNode.next.next
return True
testLinkedlist = Node(1)
testLinkedlist.createLinkedListFromList([2, 3, 4, 5])
targetNode = getNode(testLinkedlist, 3)
better_del_middle_node(targetNode)
testLinkedlist.printLinkedList()
testLinkedlist2 = Node(1)
testLinkedlist2.createLinkedListFromList([2, 3, 9, 5])
targetNode2 = getNode(testLinkedlist2, 9)
better_del_middle_node(targetNode2)
testLinkedlist2.printLinkedList()
