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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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970 Discussions

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  • abdelrahman_sam9
     + 0 comments

    function getNode(head, positionFromTail) {

    const dict = {} let i=0 while(head){ dict[i] = head.data; i++ head=head.next } i=i-1; return dict[i-positionFromTail]

    }

  • 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;
}

  • mzohaib01_inbox
     + 0 comments

    This is Python Code using stack.

    def getNode(llist, positionFromTail):
    head = llist
    stack = []
    
    while head:
        stack.append(head.data)
        head = head.next
        
    for i in range(positionFromTail+1):
        final = stack.pop()
    
    
    return final

  • gauravmali118
     + 0 comments
    def getNode(llist, positionFromTail):
    len1 = 0
    temp = llist
    while temp:
        len1 += 1
        temp = temp.next
    temp = llist
    for i in range(len1 - positionFromTail - 1):
        temp = temp.next
    return temp.data

  • nandaditra56
     + 0 comments

    This is my solution

    let arr = []

    while(llist!= null) {
    arr.push(llist.data)
    llist = llist.next
}

while(positionFromTail > 0) {
    arr.pop()
    positionFromTail--
}

return arr[arr.length-1]