Print the last k nodes of the linked list in reverse order Iterative approach in C language

We have to print the k number of nodes of the linked list in reverse order. We have to apply the iterative approach to solve this problem.

Iterative method is the one which generally uses loops that are executed till the condition holds value 1 or true.

Let’s say, list contains the nodes 29, 34, 43, 56 and 88 and the value of k is 2 than the output will be the alternate nodes till k such as 56 and 88.


Linked List: 29->34->43->56->88
Input: 2
Output: 56 88

Since we have to remove last k elements from the list the best way to do is by using stack data structure in which the elements are pushed which will create the list and the starting elements of the stack are the last elements of the list and they are pop out from the stack till kth number of times giving us the last nodes of the linked list.

The below code shows the c implementation of the algorithm given.


   Step 1 -> create node variable of type structure
      Declare int data
      Declare pointer of type node using *next
   Step 2 -> create struct node* intoList(int data)
      Create newnode using malloc
      Set newnode->data = data
      newnode->next = NULL
      return newnode
   step 3 -> Declare function void rev(struct node* head,int count, int k)
      create struct node* temp1 = head
      Loop While(temp1 != NULL)
         temp1 = temp1->next
      Declare int array[count], temp2 = count,i
      Set temp1 = head
      Loop While(temp1 != NULL)
         Set array[--temp2] = temp1->data
         Set temp1 = temp1->next
      Loop For i = 0 and i < k and i++
         Print array[i]
   Step 4 -> In Main()
      Create list using struct node* head = intoList(9)
      Set k=3 and count=0
      Call rev(head,count,k)


// Structure of a node
struct node {
   int data;
   struct node *next;
//functon for inserting a new node
struct node* intoList(int data) {
   struct node* newnode = (struct node*)malloc(sizeof(struct node));
   newnode->data = data;
   newnode->next = NULL;
   return newnode;
// Function to reversely printing the elements of a node
void rev(struct node* head,int count, int k) {
   struct node* temp1 = head;
   while(temp1 != NULL) {
      temp1 = temp1->next;
   int array[count], temp2 = count,i;
   temp1 = head;
   while(temp1 != NULL) {
      array[--temp2] = temp1->data;
      temp1 = temp1->next;
   for(i = 0; i < k; i++)
   printf("%d ",array[i]);
int main() {
reverse of a list is : ");    struct node* head = intoList(9); //inserting elements into a list    head->next = intoList(76);    head->next->next = intoList(13);    head->next->next->next = intoList(24);    head->next->next->next->next = intoList(55);    head->next->next->next->next->next = intoList(109);    int k = 3, count = 0;    rev(head, count, k); //calling function to print reversely    return 0; }


If we run above program then it will generate following output.

reverse of a list is : 109 55 24

Updated on: 22-Aug-2019


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