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.
- Make second last as last (secLast->next = NULL).
- Set next of last as head (last->next = *head_ref).
- Make last as head ( *head_ref = last).
Python3
# Python3 code to move the last item # to frontclass Node:    def __init__(self, data):        self.data = data        self.next = NoneÂ
class LinkedList:Â Â Â Â def __init__(self):Â Â Â Â Â Â Â Â self.head = NoneÂ
    # 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()Â
    # Function to bring the last node     # to the front    def moveToFront(self):        tmp = self.headÂ
        # To maintain the track of        # the second last node        sec_last = NoneÂ
        # To check whether we have not         # received the empty list or list         # with a single node        if not tmp or not tmp.next:             returnÂ
        # Iterate till the end to get        # the last and second last node         while tmp and tmp.next :            sec_last = tmp            tmp = tmp.nextÂ
        # Point the next of the second        # last node to None        sec_last.next = NoneÂ
        # Make the last node as the         # first Node        tmp.next = self.head        self.head = tmpÂ
# Driver Codeif __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!
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