Reversing a singly linked list

Singly Linked List : Reverse Operation

Idea to reverse a singly linked list is a below.

a) If there is just a one node in a singly linked list, we return it as it is as there aren’t any links to be reversed.
b) If there are more than one nodes in a singly linked list, we do the following..

  • Start with the Head node and use two additional nodes current_node and next_to_current.
    current_node stores the address of Head node.
    next_to_current stores the address of Head’s next node.


  • Point Head’s next to Null.
  • Store the address of the node next to next_to_current into a temp node before reversing the link of next_to_current.
  • Point next_to_current to current thus reversing the links between them.
  • current takes the place of next_to_current & next_to_current takes place of temp as the reversal of the links happen between every pair of adjacent nodes.


  • Repeat the link reversal steps between current and next_to_current till next_to_current points to null indicating the end of the list.


  • Thus we have the Reversed Singly linked list.


Time Complexity : O(N), where N is the number of nodes in the linked list.
Space Complexity : O(N), where N is the number of nodes in the linked list.

C++ : Reverse a singly linked list.


using namespace std;

class Node {

    int data;
    Node * next;
    Node(int arg_data) : data(arg_data), next(NULL)

class SingleLinkedList {



    void Display(Node * head) {

        Node * current = head;
        while (current != NULL) {
            cout << current->data << " ";
            current = current->next;

    Node * Reverse(Node * head) {

        // If head is null or only head (i.e single node), return head.
        if (head == NULL or head->next == NULL)
            return head;

        Node * current = head;
        Node * next_to_current = head->next;
        head->next = NULL;

        while(next_to_current != NULL) {

            Node * temp = next_to_current->next;
            next_to_current->next = current;

            current = next_to_current;
            next_to_current = temp;
        return current;

int main() {

   Node * head = new Node(2);
   head->next = new Node(3);
   head->next->next = new Node(5);
   head->next->next->next = new Node(7);
   head->next->next->next->next = new Node(11);
   head->next->next->next->next->next = new Node(13);

   SingleLinkedList s;

   cout <<"\nSingle linked list : " << endl;
   cout <<"\nReversing the single linked list : " << endl;
   head = s.Reverse(head);

   delete head->next->next->next->next->next;
   delete head->next->next->next->next;
   delete head->next->next->next;
   delete head->next->next;
   delete head->next;
   delete head;

   return 0;


Single linked list : 
2 3 5 7 11 13 
Reversing the single linked list : 
13 11 7 5 3 2

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