def test_create_partition():
list1 = LinkedList()
list1.addNode(3)
list1.addNode(1)
list1.addNode(2)
create_partition(list1, 2)
assert list1.head.value == 1
assert list1.head.next.value == 2
assert list1.head.next.next.value == 3
def test_remove_dups():
list1 = LinkedList()
list1.addNode("peanut")
list1.addNode("davey")
list1.addNode("dog")
list1.addNode("peanut")
list1.addNode("hearts")
list1.addNode("dog")
remove_dups(list1)
assert list1.head.next.next.next.value == "hearts"
from linked_list import LinkedList
# import linked_list will get all top level functionality, then you gotta do linked_list.LinkedList
import linked_list
list = LinkedList()
list.addNode(5)
list.addNode(4)
list.addNode(3)
list.addNode(5)
list.printList()
# given an unsorted linked list, remove the duplicates
# eg: 5->4->3->5
# after remove dup: 5->4->3
# as we go through the list, keep track of numbers seen in a dictionary (hashtable)
# note if the data were chars, we could potentially just have an array
def removeDups(list):
curr = list.head
elements = {}
# no special case for head element, can't have dups with only 1 element
elements[curr.data] = True
prev = curr
curr = curr.next
while curr != None:
key = curr.data
if key in elements:
prev.next = curr.next # removes curr
else:
from linked_list import LinkedList # find kth to last element of a singly linked list # Second to last = 2nd to last # 1st to last = last list = LinkedList() list.addNode(1) list.addNode(2) list.addNode(3) list.addNode(4) list.addNode(5) list.addNode(6) list.printList() # we can run through the linked list to grab the size, N. # Linked list elements are indexed 0, 1, ... N-1 # kth to last element. # 1st to last element = last element. = N - 1 # 2nd to last element = N - 2 # Nth to last element = N - N = 0 = first element def get_kth_to_last(list, k): curr = list.head num_elements = 0 while curr != None: curr = curr.next num_elements+=1 index_we_want = num_elements - k if index_we_want < 0: return None
ll.head = previousnode
#recursive solution
def recReverse(ll, node):
if node.next == None:
ll.head = node
return
recReverse(ll, node.next)
temp = node.next
temp.next = node
node.next = None
if __name__ == "__main__":
ll = LinkedList()
nodes = [1,2,3,4,5,6]
for i in nodes:
node = Node(i)
ll.addNode(node)
print(ll.print_list())
iterReverse(ll)
print(ll.print_list())
recReverse(ll, ll.head)
print(ll.print_list())
