8
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Description:

You’re given the pointer to the head node of a linked list and a specific position. Counting backwards from the tail node of the linked list, get the value of the node at the given position. A position of 0 corresponds to the tail, 1 corresponds to the node before the tail and so on.

Code:

int GetNode(Node head,int n) {
     // This is a "method-only" submission. 
     // You only need to complete this method. 
    Node current = head;
    int count = 0;
    while (current != null) {
        count++;
        current = current.next;
    }
    current = head;
    for (int i = 0; i < count - n - 1; i++) { // extra -1 to avoid going out of linked list.
        current = current.next;
    }
    return current.data;
}
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  • \$\begingroup\$ What if you are passed an empty list (head is null)? How do you want to handle the case of n >= the length of the list? Your code returns the first list element, @janos code will simply fail. In an assignment I would at least want to see that you realized there are potential problems. \$\endgroup\$ – linac Dec 23 '16 at 11:57
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Instead of a count variable and reusing current, I think it's neater to use two pointers:

Node runner = head;
for (int i = 0; i < n; i++) {
    runner = runner.next;
}
Node current = head;
while (runner.next != null) {
    runner = runner.next;
    current = current.next;
}
return current.data;

Other than that, your implementation is fine, except for a few tiny style points that are barely worth mentioning, but here we go anyway:

  • The commented out instructions about method-only submissions are unnecessary
  • Add a space after commas in parameter list
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  • \$\begingroup\$ I have to agree, a very neat solution indeed. I wonder how to get thinking like that :) \$\endgroup\$ – CodeYogi Dec 23 '16 at 6:33
  • \$\begingroup\$ Me too :-) I saw it once and can't forget about it. \$\endgroup\$ – janos Dec 23 '16 at 6:34
  • \$\begingroup\$ Implementation in C, Java and Python: geeksforgeeks.org/nth-node-from-the-end-of-a-linked-list \$\endgroup\$ – Frode Akselsen Dec 23 '16 at 6:43

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