How to find nth element from the end of a singly linked list?

Question

The following function is trying to find the nth to last element of a singly linked list.

For example:

If the elements are

Answer 1

Since this sounds like homework, I prefer to help you help yourself instead of giving an actual solution.

I suggest you run this code on some small sample dataset. Use your debugger to run lines step-by-step (you can set a breakpoint at the start of the function). This should give you an idea of how the code works.

You can also Console.WriteLine() to output variables of interest.

Answer 2

public int nthFromLast(int n){
    Node current = head;
    Node reference = head;      
    for(int i=0;i<n;i++){
        reference=reference.getNext();
    }
    while(reference != null){
        current = current.getNext();
        reference = reference.getNext();
    }
    return current.getData();
}

Answer 3

In java i will use-

public class LL {
  Node head;
  int linksCount;

   LL(){
     head = new Node();
     linksCount = 0;
   }

  //TRAVERSE TO INDEX
  public Node getNodeAt(int index){
    Node temp= head;
    if(index > linksCount){
        System.out.println("index out of bound !");
        return null;
    }
    for(int i=0;i<index && (temp.getNext() != null);i++){
        temp = temp.getNext();
    }
    return temp.getNext();
  }
}

Answer 4

To understand this problem, we should do a simple analogy with a measurement example. Let's say, you have to find the place of your arm where exactly 1 meter away from your middle finger, how would you measure? You would just grab a ruler with a 1-meter length and put the top-end of that ruler to the tip of your middle-finger and the bottom-end of the meter will be exactly 1 meter away from the top of your middle-finger.

What we do in this example will be the same, we just need a frame with n-element wide and what we have to do is put the frame to the end of the list, thus the start node of the frame will be exactly n-th element to the end of the list.

This is our list assuming we have M elements in the list, and our frame with N element wide;

HEAD -> EL(1) -> EL(2) -> ... -> EL(M-1) -> EL(M)

<-- Frame -->

However, we only need the boundaries of the frame, thus the end boundary of the frame will exactly (N-1) elements away from the start boundary of the frame. So have to only store these boundary elements. Let's call them A and B;

HEAD -> EL(1) -> EL(2) -> ... -> EL(M-1) -> EL(M)

A <- N-Element Wide-> B

The first thing we have to do is finding B, which is the end of the frame.

ListNode<T> b = head;
int count = 1;

while(count < n && b != null) {
    b = b.next;
    count++;
}

Now b is the n-th element of the array, and a is located on the HEAD. So our frame is set, what we will do is increment both boundary nodes step by step until b reachs to the end of the list where a will be n-th-to-the-last element;

ListNode<T> a = head;

while(b.next != null) {
    a = a.next;
    b = b.next;
}

return a;

To gather up everything, and with the HEAD checks, N < M (where M is the size of the list) checks and other stuff, here is the complete solution method;

public ListNode<T> findNthToLast(int n) {
    if(head == null) {
        return null;
    } else {
        ListNode<T> b = head;
        int count = 1;

        while(count < n && b != null) {
            b = b.next;
            count++;
        }

        if(count == n && b!=null) {
            ListNode<T> a = head;

            while(b.next != null) {
                a = a.next;
                b = b.next;
            }

            return a;
        } else {
            System.out.print("N(" + n + ") must be equal or smaller then the size of the list");
            return null;
        }
    }
}

Answer 5

Here is the code using 2 pointer approach : ( source )

Slow and Faster pointer approach

struct node
{
  int data;
  struct node *next;
}mynode;


mynode * nthNodeFrmEnd(mynode *head, int n /*pass 0 for last node*/)
{
  mynode *ptr1,*ptr2;
  int count;

  if(!head)
  {
    return(NULL);
  }

  ptr1  = head;
  ptr2  = head;
  count = 0;

  while(count < n)
  {
     count++;
     if((ptr1=ptr1->next)==NULL)
     {
        //Length of the linked list less than n. Error.
        return(NULL);
     }
  }

  while((ptr1=ptr1->next)!=NULL)
  {
    ptr2=ptr2->next;
  }

  return(ptr2);
}

Recursion

node* findNthNode (node* head, int find, int& found){
    if(!head) {
        found = 1;
        return 0;
    }
    node* retval = findNthNode(head->next, find, found);
    if(found==find)
        retval = head;
    found = found + 1;
    return retval;
}

---

Answer 6

Here is C# version of finding nth child from Linklist.

public Node GetNthLast(Node head, int n)
    {
        Node current, nth;
        current = nth = head;
        int counter = 0;

        while (current.next != null)
        {
            counter++;
            if (counter % n == 0)
            {
                for (var i = 0; i < n - 1; i++)
                {
                    nth = nth.next;
                }
            }
            current = current.next;
        }
        var remainingCounts = counter % n;
        for (var i = 0; i < remainingCounts; i++)
        {
            nth = nth.next;
        }
        return nth;
    }