def pad_list(h1, h2):
c1, c2 = h1, h2
l1, l2 = 0,0
while c1 != None:
l1 += 1
c1 = c1.next
while c2 != None:
l2 += 1
c2 = c2.next
if l1 < l2:
h = Node(0)
padding = [0 for _ in xrange(l1 - l2 - 1)]
h.append_array(padding)
h.append_node(h1)
return h, h2
elif l1 > l2:
h = Node(0)
padding = [0 for _ in xrange(l2 - l1 - 1)]
h.append_array(padding)
h.append_node(h2)
return h1, h
else:
return h1, h2
"""
def remove_duplicate(head):
dups = {}
curr, prev = head, None
while curr != None:
if dups.get(curr.data, False) == False:
dups[curr.data] = True
prev = curr
else:
prev.next = curr.next
curr = curr.next
return head
head = Node(5)
head.append_array([2, 3, 4, 2, 3, 4])
print "remove duplicate [5, 2, 3, 4, 2, 3, 4]", remove_duplicate(head).to_list() == [5, 2, 3, 4]
""" Return kth to last
Implement an algorithm to find the kth to last element of a singly linked list
"""
def kth_to_last(head, k):
if k == 0:
return Node(None)
first, second = head, head
while k > 0:
k -= 1
first = first.next
while first != None:
first, second = first.next, second.next