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  1. Prepare
  2. Data Structures
  3. Linked Lists
  4. Get Node Value
  5. Discussions

Get Node Value

  • oneton
     + 0 comments

    I see a lot of solutions posted with double loops and/or a temporary stack of values. It's quite easy to save on repeats and memory with a single pass through the list and keeping track of 2 pointers. Give the one "searching for the end" a "head start" of the amount of provided positions and when you've reached the number, update the pointer to the node with the data.

    public static int getNode(SinglyLinkedListNode llist, int positionFromTail) {
    SinglyLinkedListNode dataNode = llist;
    SinglyLinkedListNode current = llist;
    int ahead = 0;
    while (current.next != null) {
        if (ahead == positionFromTail) {
            dataNode = dataNode.next;
        } else {
            ahead++;
        }

        current = current.next;
    }

    return dataNode.data;
}