C++ Program For Moving Last Element To Front Of A Given Linked List Last Updated : 03 Aug, 2022 Comments Improve Suggest changes Like Article Like Report 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). C++ /* C++ Program to move last element to front in a given linked list */ #include <bits/stdc++.h> using namespace std; // A linked list node class Node { public: int data; Node *next; }; /* We are using a double pointer head_ref here because we change head of the linked list inside this function.*/ void moveToFront(Node **head_ref) { /* If linked list is empty, or it contains only one node, then nothing needs to be done, simply return */ if (*head_ref == NULL || (*head_ref)->next == NULL) return; /* Initialize second last and last pointers */ Node *secLast = NULL; Node *last = *head_ref; /* After this loop secLast contains address of second last node and last contains address of last node in Linked List */ while (last->next != NULL) { secLast = last; last = last->next; } // Set the next of second last as NULL secLast->next = NULL; // Set next of last as head node last->next = *head_ref; /* Change the head pointer to point to last node now */ *head_ref = last; } // UTILITY FUNCTIONS /* Function to add a node at the beginning of Linked List */ void push(Node** head_ref, int new_data) { // Allocate node Node* new_node = new Node(); // Put in the data new_node->data = new_data; // Link the old list off the // new node new_node->next = (*head_ref); // Move the head to point to the // new node (*head_ref) = new_node; } /* Function to print nodes in a given linked list */ void printList(Node *node) { while(node != NULL) { cout << node->data << " "; node = node->next; } } // Driver code int main() { Node *start = NULL; /* The constructed linked list is: 1->2->3->4->5 */ push(&start, 5); push(&start, 4); push(&start, 3); push(&start, 2); push(&start, 1); cout << "Linked list before moving last to front"; printList(start); moveToFront(&start); cout << "Linked list after removing last to front"; printList(start); return 0; } // This code is contributed by rathbhupendra 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. Auxiliary Space: O(1) because using constant variables Please refer complete article on Move last element to front of a given Linked List for more details! Comment More infoAdvertise with us Next Article C++ Program For Moving Last Element To Front Of A Given Linked List K kartik Follow Improve Article Tags : Linked List C++ Programs C++ DSA Linked Lists +1 More Practice Tags : CPPLinked List Similar Reads C++ Program For Deleting Last Occurrence Of An Item From Linked List Using pointers, loop through the whole list and keep track of the node prior to the node containing the last occurrence key using a special pointer. After this just store the next of next of the special pointer, into to next of special pointer to remove the required node from the linked list. 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