Swap Kth node from beginning with Kth node from end in a Linked List | GeeksforGeeks

Given a singly linked list, swap kth node from beginning with kth node from end. Swapping of data is not allowed, only pointers should be changed.
Let X be the kth node from beginning and Y be the kth node from end. Following are the interesting cases that must be handled.
1) Y is next to X
2) X is next to Y
3) X and Y are same
4) X and Y don’t exist (k is more than number of nodes in 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 beginning of 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;

}

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