def BackwordSum(ll1, ll2):
s1 = StringifyList(ll1)
s2 = StringifyList(ll2)
s3 = str(int(s1[::-1]) + int(s2[::-1]))
ll3 = LinkedList()
for i in s3[::-1]:
ll3.appendNode(i)
def ForwordSum(ll1, ll2):
s1 = StringifyList(ll1)
s2 = StringifyList(ll2)
s3 = str(int(s1) + int(s2))
ll3 = LinkedList()
for i in s3:
ll3.appendNode(i)
# Implement an algorithm to delete a node in the middle (i.e., any node but the first and last node, not necessarily the exact middle) of a singly linked list, given only access to that node.
# EXAMPLE
# lnput:the node c from the linked lista->b->c->d->e->f
# Result: nothing is returned, but the new linked list looks likea->b->d->e- >f
from linkedList import LinkedList
ll = LinkedList()
ll.appendNode(1)
ll.appendNode(2)
ll.appendNode(2)
ll.appendNode(1)
ll.appendNode(2)
ll.appendNode(3)
ll.appendNode(3)
ll.appendNode(4)
ll.appendNode(4)
ll.appendNode(5)
ll.appendNode(5)
def deleteNode(ll, value):
if ll.head.value == value:
oldHead = ll.head
ll.head = ll.head.next
oldHead.next = None
else:
node = ll.head
while node and node.next:
if node.next.value == value:
nextNode = node.next
node.next = nextNode.next
# Write code to partition a linked list around a value x, such that all nodes less than x come before all nodes greater than or equal to x. If x is contained within the list, the values of x only need to be after the elements less than x (see below). The partition element x can appear anywhere in the "right partition"; it does not need to appear between the left and right partitions.
# EXAMPLE
# Input: 3 -> 5 -> 8 -> 5 -> 10 -> 2 -> 1 [partition=5]
# Output: 3 -> 1 -> 2 -> 10 -> 5 -> 5 -> 8
from linkedList import LinkedList
ll = LinkedList()
ll.appendNode(11)
ll.appendNode(3)
ll.appendNode(8)
ll.appendNode(5)
ll.appendNode(10)
ll.appendNode(2)
ll.appendNode(12)
def Partition(ll, value):
node = ll.head.next
prev = ll.head
dict = {}
while node:
if node.value < value:
prev.next = node.next
old_head = ll.head
ll.head = node
node.next = old_head
node = prev.next
else:
prev = node
node = node.next
# Given two (singly) linked lists, determine if the two lists intersect. Return the inter secting node. Note that the intersection is defined based on reference, not value.That is, if the kth node of the first linked list is the exact same node (by reference) as the jth node of the second linked list, then they are intersecting.
from linkedList import LinkedList
ll1 = LinkedList()
ll1.appendNode(11)
ll1.appendNode(3)
ll1.appendNode(8)
ll1.appendNode(5)
ll1.appendNode(10)
ll1.appendNode(2)
ll1.appendNode(12)
ll2 = LinkedList()
ll2.appendNode(11)
ll2.appendNode(3)
ll2.appendNode(8)
ll2.appendNode(5)
ll2.appendNode(10)
ll2.appendNode(2)
ll2.appendNode(12)
def CheckIntersection(ll1, ll2):
node1 = ll1.head
node2 = ll2.head
dict = {}
while node1:
dict[node1.value] = node1.next
node1 = node1.next
while node2:
