def test_intersects__yes():
left = LinkedList([1, 2, 3, 4, 5])
element = left[3]
right = LinkedList([1, 2])
right.append(element)
assert intersects(left, right) == element
def partition(ll, p):
"""
Admittedly, much of the implementation magic here is making sure that the
basic dunder method __add__ to join two linked lists works correctly. I
also rewrote the LinkedList initializer slightly to make use of a new
append method that appends an Element to the end of a LinkedList and
adjusts the prev/next pointers and the tail pointer correctly.
"""
left = LinkedList([])
right = LinkedList([])
for value in ll.values:
element = Element(value)
if value < p:
left.append(element)
else:
right.append(element)
return left + right
def sum(left, right, reverse=True):
"""
Summation of digits may "carry over" a value to the next digit place.
When evaluating the linked list, it's simple to just take the carry along
the traversal of both left and right in tandem, and apply it to the next
digit place. If one list terminates before the other, the carry gets
applied just the same until the traversal ends for the longer list. This
algorithm runs in O(M+N) time where M, N is the length of left, right
respectively. The sum of each individual element can be applied into a new
element and just appended to a new linked list.
The non-reversed case is a bit trickier, and requires special handling for
the possibility of non-equal list lengths, as we can't assume the first
elements are the same digit place. Once we determine the maximum length,
we can use a hash to store each digit place keyed by power of 10. Using
the hash allows us to keep the runtime to O(M+N) at the cost of
O(max(M, N)) space. Other methods involve complicated back-tracking with
pointers, or using a stack which is essentially the first method.
The non-reserved solution avoids fun workarounds like reversing the linked
list and running the first algorithm, or converting the LinkedList into
numbers, summing those, and then converting the number into a string and
then passing it directly into LinkedList... :rolling_on_the_floor_laughing:
"""
result = LinkedList([])
if reverse:
current_left = left.head
current_right = right.head
carry = 0
while current_left is not None and current_right is not None:
left_value = 0
if current_left is not None:
left_value = current_left.value
current_left = current_left.next
right_value = 0
if current_right is not None:
right_value = current_right.value
current_right = current_right.next
total = left_value + right_value + carry
carry = total // 10
digit = total % 10
result.append(Element(digit))
if carry > 0:
result.append(Element(carry))
else:
# this is needed here in case there is carry past the max digit place
places = defaultdict(lambda: 0)
for ll in [left, right]:
for index, element in enumerate(ll):
place = len(ll) - index - 1
places[place] += element.value
for place in sorted(places.keys()):
carry = places[place] // 10
places[place] %= 10
if carry > 0:
places[place + 1] += carry
for place in sorted(places.keys(), reverse=True):
result.append(Element(places[place]))
return result