Move last element to front of a given linked list Practice

Move last element to front of a Linked-List in C++

In this tutorial, we will learn to move last element to front of a linked-list in C++.

For Example:

• If the given linked list is 1->2->3->4->5
• Then, the output is 5->1->2->3->4

Algorithm:

• Firstly, traverse the linked list till the last node.
• Secondly, we will use two pointers here.
• Now, use one pointer to store the address of the last node and second pointer to store the address of the second last node.
• Do following after the end of the loops.
• Now make the second last node as the last node.
• Then set next of the last node as head of the list.
• Finally, make the last node as the head of the list.
• The time complexity of the solution is O[s] where s is the total number of nodes in the list.

You may also like:

• Singly Linked Lists in C++
• Doubly Linked Lists in C++

Program to move the last element to the front of a given Linked-List in C++

Hence, the implementation is here.

Code:

#include #include using namespace std; // node class class Node { public: int dat; Node *nex; }; void mvefront[Node **head] { if [*head == NULL || [*head]->nex == NULL] return; Node *seclast = NULL; Node *last = *head; while [last->nex != NULL] { seclast = last; last = last->nex; } seclast->nex = NULL; last->nex = *head; *head = last; } void push[Node** head, int newdata] { Node* newnode = new Node[]; newnode->dat = newdata; newnode->nex = [*head]; [*head] = newnode; } //print void printl[Node *node] { while[node != NULL] { cout dat nex; } } int main[] { Node *strt = NULL; push[&strt, 5]; push[&strt, 4]; push[&strt, 3]; push[&strt, 2]; push[&strt, 1]; printl[strt]; cout