Let last be the first

Level EASY

Write a function that moves the last element to the front in a given singly Linked List.

For example, if the given Linked List is 1->2->3->4->5, then the function should change the list to 5->1->2->3->4.

You just need to return the head of the new linked list, don't print the elements.

Input format :

Line 1 : Linked list elements of length n (separated by space and terminated by -1)

Output format :

Updated list elements (separated by space)

Constraints :

1 <= n <= 10^4

Sample Input :

 1 2 3 4 5 6 -1

Note: -1 at the end of input is just a terminator representing the end of linked list. This -1 is not part of the linked list. Size of given linked list is 6.

Sample Output :

 6 1 2 3 4 5

	## Read input as specified in the question.
	## Print output as specified in the question.
	class Node:
	def __init__(self,data):
	self.data=data
	self.next=None
	def LinkedLists(arr):
	head=None
	tail=None
	if len(arr)<1:
	return -1
	else:
	for i in arr:
	if i == -1:
	break
	NewNode = Node(i)
	if head==None:
	head=NewNode
	tail=NewNode
	else:
	tail.next=NewNode
	tail=NewNode
	return head
	def appendLastNToFirst(head,k):
	h0=head
	total = 1
	while ( h0 is not None):
	h0 = h0.next
	total += 1
	current = head
	count = 1
	while (count < total-k-1 and current is not None):
	current = current.next
	count += 1
	NewNode=current
	while (current.next is not None):
	current = current.next
	current.next = head
	head = NewNode.next
	NewNode.next = None
	printLL(head)
	def printLL(head):
	while head is not None:
	print(head.data,end=" ")
	head=head.next
	arr=list(map(int,input().split()))
	appendLastNToFirst(LinkedLists(arr),1)

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