GeeksforGeeks App
Open App
Browser
Continue

# C++ 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).

## C++

 `/* C++ Program to move last element``   ``to front in a given linked list */``#include ``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!

My Personal Notes arrow_drop_up