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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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973 Discussions

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  • akhileshpra_sha1
     + 1 comment
    def atNodeFromTail(head,position):
    cur = head
    stack = []
    while cur != None:
        stack.append(cur.data)
        cur = cur.next
    return stack[-position-1]

    Simple Solution with O(N) complexity

  • pierbus_dev
     + 0 comments

    typescript, just recursion, not creating a stack or a dict of the visited nodes

    function traverse (llist: SinglyLinkedListNode, positionFromTail: number, found: { value: number}): number {
    if(llist.next !== null){
        const pos = 1 + traverse(llist.next, positionFromTail, found)
        if (pos === positionFromTail){
            found.value = llist.data
        }
        return pos
    }
    if (positionFromTail === 0){
        found.value = llist.data
    }
    return 0
}

function getNode(llist: SinglyLinkedListNode, positionFromTail: number): number {
    const found = {  value: 0}
    traverse(llist, positionFromTail, found)
    return found.value
}

  • nicksonkipkorir1
     + 0 comments
        Stack<Integer> stack = new Stack<>();

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

for (int i = 0; i < positionFromTail; i++) {
    stack.pop();
}

// The top of the stack is now the desired node
return stack.pop();

  • 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;
}