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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