C++ Program For Swapping Kth Node From Beginning With Kth Node From End In A Linked List

// A C++ program to swap Kth node
// from beginning with kth node from end
#include <bits/stdc++.h>
using namespace std;

// A Linked List node
struct Node 
{
    int data;
    struct Node* next;
};

/* Utility function to insert
   a node at the beginning */
void push(struct Node** head_ref, 
          int new_data)
{
    struct Node* new_node = 
    (struct Node*)malloc(sizeof(struct Node));
    new_node->data = new_data;
    new_node->next = (*head_ref);
    (*head_ref) = new_node;
}

/* Utility function for displaying 
   linked list */
void printList(struct Node* node)
{
    while (node != NULL) 
    {
        cout << node->data << " ";
        node = node->next;
    }
    cout << endl;
}

/* Utility function for calculating 
   length of linked list */
int countNodes(struct Node* s)
{
    int count = 0;
    while (s != NULL) 
    {
        count++;
        s = s->next;
    }
    return count;
}

/* Function for swapping kth nodes 
   from both ends of linked list */
void swapKth(struct Node** head_ref, int k)
{
    // Count nodes in linked list
    int n = countNodes(*head_ref);

    // Check if k is valid
    if (n < k)
        return;

    // If x (kth node from start) and
    // y(kth node from end) are same
    if (2 * k - 1 == n)
        return;

    // Find the kth node from the beginning 
    // of the linked list. We also find
    // previous of kth node because we
    // need to update next pointer of
    // the previous.
    Node* x = *head_ref;
    Node* x_prev = NULL;
    for (int i = 1; i < k; i++) 
    {
        x_prev = x;
        x = x->next;
    }

    // Similarly, find the kth node from
    // end and its previous. kth node
    // from end is (n-k+1)th node from 
    // beginning
    Node* y = *head_ref;
    Node* y_prev = NULL;
    for (int i = 1; i < n - k + 1; i++) 
    {
        y_prev = y;
        y = y->next;
    }

    // If x_prev exists, then new next of
    // it will be y. Consider the case
    // when y->next is x, in this case,
    // x_prev and y are same. So the statement
    // "x_prev->next = y" creates a self loop.
    // This self loop will be broken
    // when we change y->next.
    if (x_prev)
        x_prev->next = y;

    // Same thing applies to y_prev
    if (y_prev)
        y_prev->next = x;

    // Swap next pointers of x and y.
    // These statements also break self
    // loop if x->next is y or y->next is x
    Node* temp = x->next;
    x->next = y->next;
    y->next = temp;

    // Change head pointers when k is 1 or n
    if (k == 1)
        *head_ref = y;
    if (k == n)
        *head_ref = x;
}

// Driver code
int main()
{
    // Let us create the following
    // linked list for testing
    // 1->2->3->4->5->6->7->8
    struct Node* head = NULL;
    for (int i = 8; i >= 1; i--)
        push(&head, i);

    cout << "Original Linked List: ";
    printList(head);

    for (int k = 1; k < 9; k++) 
    {
        swapKth(&head, k);
        cout << 
        "Modified List for k = " << k << endl;
        printList(head);
    }

    return 0;
}