def reverse_ll_2(ll):
return reverse(ll, None)
def reverse(node, prev_node):
if node is None:
# the end of the ll is reached, return the previous node
# that'll be the first node in the reversed ll
return prev_node
# send node.next as current node and node as previous node in the next step
result = reverse(node.next, node)
# change the pointer of the current node to point to the previous node
node.next = prev_node
return result
###########
# Testing #
###########
# import build_ll and print_ll methods from ll_helpers.py
from ll_helpers import build_ll, print_ll
# Test 1
# Correct result => 4 -> 3 -> 2 -> 1
print_ll(reverse_ll(build_ll([1, 2, 3, 4])))
print_ll(reverse_ll_2(build_ll([1, 2, 3, 4])))
total = transfer + v1 + v2
transfer = total // 10
if l1 is None:
# if the first list is shorter than the second, add new elements at the end
pointer.next = ListNode(None)
pointer = pointer.next
pointer.val = total % 10
return start.next
###########
# Testing #
###########
# import build_ll and print_ll methods from ll_helpers.py
from ll_helpers import build_ll, print_ll
# Test 1
# Correct result => 7 -> 0 -> 8
ll1 = build_ll([2, 4, 3])
ll2 = build_ll([5, 6, 4])
print_ll(add_two_numbers(ll1, ll2))
# Test 2
# Correct result => 8 -> 9 -> 0 -> 0 -> 1
ll1 = build_ll([9, 9, 9, 9])
ll2 = build_ll([9, 9])
print_ll(add_two_numbers(ll1, ll2))
def remove_duplicates(nums):
if nums is None:
return nums
pointer = nums
while pointer.next is not None:
if pointer.val == pointer.next.val:
# skip the next value because it's a duplicate
pointer.next = pointer.next.next
else:
# search next
pointer = pointer.next
return nums
###########
# Testing #
###########
# import build_ll and print_ll methods from ll_helpers.py
from ll_helpers import build_ll, print_ll
# Test 1
# Correct result => 1 -> 2
print_ll(remove_duplicates(build_ll([1, 1, 2])))
# Test 2
# Correct result => 0 -> 1 -> 2 -> 3 -> 4
print_ll(remove_duplicates(build_ll([0, 0, 1, 1, 1, 2, 2, 3, 3, 4])))
while head is not None:
if i % 2 == 1:
oddPointer.next = head
oddPointer = oddPointer.next
else:
evenPointer.next = head
evenPointer = evenPointer.next
head = head.next
i += 1
evenPointer.next = None
oddPointer.next = even.next
return odd.next
###########
# Testing #
###########
# import build_ll and print_ll methods from ll_helpers.py
from ll_helpers import build_ll, print_ll
# Test 1
# Correct result => 1 -> 3 -> 5 -> 2 -> 4
print_ll(odd_even_ll(build_ll([1, 2, 3, 4, 5])))
# Test 2
# Correct result => 2 -> 3 -> 6 -> 7 -> 1 -> 5 -> 4
print_ll(odd_even_ll(build_ll([2, 1, 3, 5, 6, 4, 7])))