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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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975 Discussions

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  • Lonchik
     + 0 comments

    Yet another C# broken template (5th now?)

    remove result class, but keep its guts. make method non-public.

  • ouadochhassan201
     + 0 comments

    solution with JS

    function getNode(llist, positionFromTail) {
    // Write your code here
    var currentNode = llist;
    var len = 0
    var arr = [];
    if(llist==null){
        return false;
    }
    while(currentNode){
        len++;
        arr.unshift(currentNode.data)
        currentNode = currentNode.next
    }
        if(positionFromTail>len || positionFromTail<0){
            return false
        }

    return arr[positionFromTail]
    
}

  • 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();