CareerCup

traverse the linked list and, when the kth element is found, fix the next-pointer of the previous and next element. of course you have to take care of the corner cases such as when there are less than k elements.

Put two pointers p and q to head.
Traverse with q to k places.Keep p at head.
Move p and q forward,from their respective postitions till q reaches end.
link pointed by p is the required node.

keep one pointer at first element and another pointer at k th element from first and increment the pointers till second pointer reaches null ,then the first pointer is at k th element from last ,so you and find it in this way..........

I assume it should be done in a single-pass

// assuming k is valid
ListNode* deleteKthNodeFromBack(ListNode* head, int k)
{
    if (head == NULL) return head;
    ListNode* fast (head), slow (head) slowPrev (NULL);
    while (--k) fast = fast->next;
    while (fast->next != NULL) {
          slowPrev = slow;
          slow = slow->next;
          fast = fast->next;
    }
    if (slowPrev == NULL) head = slow->next;
    else slowPrev->next = slow->next;
   
    delete slow;
    return head;
}

public static Elem head; 

 public static int DeleteKth(Elem elem,Elem prev,int k) {
            int count = 0;
            if (elem != null) {
                count = DeleteKth(elem.sled,elem, k)+1;
                if (count == k) {
                    if (prev == null) head = elem.sled;
                    else prev.sled = elem.sled;
                }
            }        
            return count;
        }
// call with 
 DeleteKth(head, null, 2); where 2 is kth to be deleted

Just a few name corrections

public static int DeleteKth(Elem elem,Elem prev,int k) {
            int count = 0;
            if (elem != null) {
                count = DeleteKth(elem.next,elem,k)+1;
                if (count == k) {
                    if (prev == null) prvi = elem.next;
                    else prev.next = elem.next;
                }
            }        
            return count;
        }