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

Get Node Value

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984 Discussions

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  • AndyScript
     + 0 comments

    JavaScript

    function getNode(llist, positionFromTail) {
  const values = [];
   
  let currentNode = llist;
  
  while (currentNode) {   
    values.push(currentNode.data);
    
    currentNode = currentNode.next;
  }
  
  return values.reverse()[positionFromTail];
}

  • ayoadeleke2
     + 0 comments

    Javascript solution 

    function getNode(llist, tailPos) {
    let curPos = llist

    while(curPos!= null){
        tailPos--
        curPos = curPos.next
    }
    
    tailPos = ((tailPos*-1) - 1)
    
    while(tailPos != 0){
        llist = llist.next
        tailPos--
    }
    return llist.data


}

  • nizmu
     + 0 comments

    My Python Soluction

    def getNode(llist, positionFromTail):

        prev = None
        curr = llist
        temp = curr

        while curr:
                temp = curr.next
                curr.next = prev
                prev = curr
                curr = temp

        head = prev
        curr = head
        count = 0

        while count < positionFromTail:
                curr = curr.next
                count += 1

        return curr.data

  • mdsanz
     + 0 comments

    My Java 8 Solution

    public static int getNode(SinglyLinkedListNode llist, int positionFromTail) {
        int lenght = 0;
        SinglyLinkedListNode counterList = llist;
        
        while (counterList != null) {
            lenght++;
            counterList = counterList.next;
        }
        
        int positionFromHead = lenght - positionFromTail - 1;
        int index = 0;
        SinglyLinkedListNode node = llist;
        
        while (index < positionFromHead) {
            node = node.next;
            index++;
        }
        
        return node.data;
    }

  • rajpatel92111213
     + 0 comments
    public static int getNode(SinglyLinkedListNode head, int positionFromTail) {
    // Create two pointers, both starting at the head
    SinglyLinkedListNode fast = head;
    SinglyLinkedListNode slow = head;

    // Move the 'fast' pointer 'positionFromTail' steps ahead
    for (int i = 0; i < positionFromTail; i++) {
        fast = fast.next;
    }

    // Now move both 'fast' and 'slow' one step at a time
    // When 'fast' reaches the last node, 'slow' will be at the target node
    while (fast.next != null) {
        fast = fast.next;
        slow = slow.next;
    }

    // 'slow' is now pointing to the node that is 'positionFromTail' from the end
    return slow.data;
}