# 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.
from Class_Linkedlist import randomLinkedList
def deleteNode(linkedlist, node_value):
curr = linkedlist.head
while curr.next is not None:
if curr.next.value == node_value:
curr.next = curr.next.next
return
else:
curr = curr.next
print('{} value is not in list'.format(node_value))
L1 = randomLinkedList(9, 1, 8)
print(L1)
deleteNode(L1, 5)
print(L1)
while runn.next is not None:
if runn.next.value == curr.value:
runn.next = runn.next.next
else:
runn = runn.next
curr = curr.next
# Alternate solution
def delDuplicates2(linkedlist): #O(n) time, O(n) space solution
curval = linkedlist.head
tempdict = {curval.value : True}
while curval.next is not None:
if curval.next.value in tempdict:
curval.next = curval.next.next
else:
tempdict[curval.next.value] = True
curval = curval.next
L1 = randomLinkedList(9, 3, 7) #Length, min ,max
print(L1)
delDuplicates(L1)
print(L1)
L2 = randomLinkedList(9, 3, 7)
print(L2)
delDuplicates2(L2)
print(L2)
result = []
while head1 is not None or head2 is not None:
h1val = 0 if head1 is None else head1.value
h2val = 0 if head2 is None else head2.value
temp = h1val + h2val + carry
carry = 0
if temp > 9:
result.append(int(temp % 10))
carry = int(temp / 10)
else:
result.append(temp)
if head1:
head1 = head1.next
if head2:
head2 = head2.next
if carry:
result.append(carry)
return result
L1 = randomLinkedList(5, 1, 9)
print(L1)
L2 = randomLinkedList(2, 1, 9)
print(L2)
results = add_linkedlist(L1, L2)
#print(results)
res = LinkedList()
for i in results[::-1]:
res.makeNode(i)
res.printlist()
while ptr1 is not None:
ptr1 = ptr1.next
ptr2 = ptr2.next
return ptr2.value # O(n) time and O(1) space.
#Here i am not counting the length of my Linked list.
# Deleting Kth from Last
def deleteKthfromlast(linkedlist, index):
lenll = getLength(linkedlist)
if index == 0:
return "Give index greater than 0"
if index > lenll:
return '{}th index is out of bound'.format(index)
elem = lenll - index
curr = linkedlist.head
while elem > 1:
curr = curr.next
elem -= 1
curr.next = curr.next.next
L1 = randomLinkedList(9, 2, 9)
print(L1)
print("My approach: " + str(getKthfromlast(L1, 6)))
print("Iterative 2 pointer appraoch: " + str(iterativeKthElement(L1, 6)))
getKthWithRecursion(L1.head, 6)
# deleteKthfromlast(L1,6)
# print(L1)