Print reverse of a Linked List without extra space and modification in C Program.

The task is to print the nodes starting from the end of the linked list without using extra space which means there shouldn’t be any extra variable instead the head pointer pointing the first node will be moved.


Input: 10 21 33 42 89
Output: 89 42 33 21 10

There can be many solutions to print the linked list in reverse order, like recursive approach (use extra space), reverse the linked list(requires modification in the given Linked List), pushing the elements on a stack and then pop and display the elements one by one(requires space O(n)), but these solutions seem to be using more space than the O(1).

To achieve the result without using more than O(1) we can −

  • Count the number of nodes in a linked list
  • Loop from i = n to 1 and print the i-th location’s node.


Step 1 -> create node variable of type structure
   Declare int data
   Declare pointer of type node using *next
Step 2 ->Declare function int get(struct node* head)
   Declare variable as int count=0
   Declare struct node *newme=head
   Loop While newme!=NULL
      Increment count by 1
      Set newme = newme->next
   Return count
Step 3 -> Declare Function void push(node** headref, char newdata)
   Allocate memory using malloc
   Set newnode->data = newdata
   Set newnode->next = (*headref)
   Set (*headref) = newnode
Step 4 -> Declare function int getN(struct node* head, int n)
   Declare struct node* cur = head
   Loop for int i=0 and i<n-1 && cur != NULL and i++
   Set cur=cur->next
Return cur->dataStep 5 -> Declare function void reverse(node *head)
   Declare int n = get(head)
   Loop For int i=n and i>=1 and i—
      Print getN(head,i)
Step 6 ->In Main()
   Create list using node* head = NULL
   Insert elements through push(&head, 89)
   Call reverse(head)


//node structure
struct node {
   int data;
   struct node* next;
void push(struct node** headref, int newdata) {
   struct node* newnode = (struct node*) malloc(sizeof(struct node));
   newnode->data = newdata;
   newnode->next = (*headref);
   (*headref) = newnode;
int get(struct node* head) {
   int count = 0;
   struct node* newme = head;
   while (newme != NULL){
      newme = newme->next;
   return count;
int getN(struct node* head, int n) {
   struct node* cur = head;
   for (int i=0; i<n-1 && cur != NULL; i++)
      cur = cur->next;
   return cur->data;
void reverse(node *head) {
   int n = get(head);
   for (int i=n; i>=1; i--)
      printf("%d ", getN(head, i));
int main() {
   struct node* head = NULL; //create a first node
   push(&head, 89); //pushing element in the list
   push(&head, 42);
   push(&head, 33);
   push(&head, 21);
   push(&head, 10);
   reverse(head); //calling reverse function
   return 0;


if we run above program then it will generate following output

89 42 33 21 10

Updated on: 22-Aug-2019


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