Python Program For Moving Last Element To Front Of A Given Linked List

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. Algorithm: Traverse the list till the last node. Use two pointers: one to store the address of the last node and the other for the address of the second last node. After the end of the loop do the following operations.

1. Make second last as last (secLast->next = NULL).
2. Set next of last as head (last->next = *head_ref).
3. Make last as head ( *head_ref = last).

Python3

# Python3 code to move the last item
# to front
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

class LinkedList:
    def __init__(self):
        self.head = None

    # Function to add a node
    # at the beginning of Linked List
    def push(self, data):
        new_node = Node(data)
        new_node.next = self.head
        self.head = new_node

    # Function to print nodes in
    # a given linked list
    def printList(self):
        tmp = self.head
        while tmp is not None:
            print(tmp.data, end = ", ")
            tmp = tmp.next
        print()

    # Function to bring the last node
    # to the front
    def moveToFront(self):
        tmp = self.head

        # To maintain the track of
        # the second last node
        sec_last = None

        # To check whether we have not
        # received the empty list or list
        # with a single node
        if not tmp or not tmp.next:
            return

        # Iterate till the end to get
        # the last and second last node
        while tmp and tmp.next :
            sec_last = tmp
            tmp = tmp.next

        # Point the next of the second
        # last node to None
        sec_last.next = None

        # Make the last node as the
        # first Node
        tmp.next = self.head
        self.head = tmp

# Driver Code
if __name__ == '__main__':
    llist = LinkedList()

    # Swap the 2 nodes
    llist.push(5)
    llist.push(4)
    llist.push(3)
    llist.push(2)
    llist.push(1)
    print (
    "Linked List before moving last to front ")
    llist.printList()
    llist.moveToFront()
    print (
    "Linked List after moving last to front ")
    llist.printList()

Output:

Linked list before moving last to front 
1 2 3 4 5 
Linked list after removing last to front 
5 1 2 3 4

Time Complexity: O(n) where n is the number of nodes in the given Linked List.

Space Complexity: O(1) because using constant variables

Please refer complete article on Move last element to front of a given Linked List for more details!