## Use Hacker Noon's RSS Feed

Visit Hacker Noon RSS Feed hackernoon.com/feed

promotedSeptember 7th 2019 1,283 reads

I am a passionate Fullstack developer skilled in JavaScript | webpack | React | Ruby | Ruby on Rails

A list is a collection of elements. A linked List is a list in which each element in the list contains both **data **and a **pointer **to one or both neighboring items. L**inked List** is made up of elements(nodes) which are connected doubly or singly. When the nodes are connected doubly, we refer to such a list as a** doubly-linked **list. However, a singly linked list is a sequence of the element where the first node links to the second and the second links to the third and so on in one direction… On the other hand, a **singly-linked** list has nodes with pointers which always point to the next element in the list. This discussion focus on **singly-linked list. Doubly-linked list **would be discussed in my subsequent article.

Let us consider the list: **10 →20 →30 **

We can say **10 **points to **20, 20 **points to** 30**. The element of a list is referred to as a Node. So in this case 10 →20 →30, we would say **Node10** points** **to **Node20, **and** Node20 **points to **Node30. **Each node in a singly linked list has a .**next** attribute to indicate the next node after it (the current node).

Most people are used to using Arrays. The nature of arrays makes them efficient when we want to read data at a given position in the array, as this operation is run in constant time. Arrays are inefficient when it comes to insertion and deletion. Deletion and insertion are implemented efficiently with Linked List. Adding or deleting at the beginning of a list runs at constant time O(1). Adding at the given position also runs at linear time O(n). For instance, after deleting an element from an array the indexes of each element of the array would have to be readjusted because arrays are *ordered** *collections of *items** *indexed by *contiguous integers**.*

First, we would create a node to store the data. Nodes of a singly-linked list have two attributes: **value( ***value of current node***)** and **next_node(***node after current***)**

```
class Node
attr_accessor :value, :next_node
def initialize(value, next_node = nil)
@value = value
@next_node = next_node
end
end
```

Now that we have our storage let us add elements at the end of the list. This operation is an O(n) operation, to do.

For example, let us place

**15**

after 10 the list below"**Note:*** The algorithm for this would be to: create a new node. Make the node with value 10(head), point to the new node in this case the node with value 15 and make the new node with value 15 point to the node with value 20 (point head.next) which was the .next of the head before 15 was added.*"

Before beginning adding to the list, consider where in the list you intend to add to.

1. Add to an empty list(in our case list is not empty)

2. Add at the first position

3. Add at any position aside the first position (somewhere in the middle)

4. Add at the end of the list

```
#create a new node
#if the head is nil it means the list is empty
#this would imply that the new node is the head
#if head is not nil then there are element in the list
# make a copy of head
#loop through the list from the head using the copy of the head
#note when .next_node is nil current node is the last element
#add an element at the end of the list by pointing its to the new node(the node you intend to insert) instead of nil
class LinkedList
#setup head ,remmember we traverse a linkedlist from the head
def initialize
@head = nil
@tail = nil
end
def add(number)
#create a new node
this_node = Node.new(number)
if @head.nil?
@head = this_node
return
end
current = @head
#until current.nil means until we reach the last node
until current.next_node.nil?
current = current.next_node
end
#point current(last node) to our new node
current.next_node = this_node
end
end
```

Now let us consider adding an element at a specific position rather than adding at the end. The only difference between the two is that instead of looping to the end(till **current.next.nil?**), we will loop up to the position we want to insert at.

```
#if the node is empty the new node becomes the head
#to insert at 0 or first position pass a second argument to the node
#the second argument then becomes the second node and the first argument becomes the new head
#Otherwise loop to the desired position an make the insertion
class LinkedList
def add_at(index,item)
if @head.nil?
#if list is empty, the head is the new node
this_nod=Node.new(item)
@head=this_nod
end
if index==0
# if index is 0, we insert in the first position
this_nod=Node.new(item,@head)
@head=this_nod
end
if index >0
#insert at desired position if index is greater than 0
ind=index-1
current=@head
before_current=@head
#loop to the desired position before where you wish to insert
ind.times do
before_current=current.next_node
end
#loop to the desired position where you wish to insert
index.times do
current=current.next_node
end
#create a new node you wish to insert
this_nod=Node.new(item)
after_current=before_current.next_node
#point node before current to new node
before_current.next_node=this_nod
#point new node to the old current node
this_nod.next_node=after_current
end
end
end
```

let us consider deleting or removing an element from the a list. Assuming we want to delete(remove) 20 from our list

Now let us try to remove an element from the list.

```
#if list is empty head.nil? is true so return "storage is empty"
#to insert at first position make a copy of the head
#make the copy of the head .next point to the head thus making the head the second element
#insert at given position by looping to the said index make the previous node point to the .next value of the current node
#make the current node point to itself
class LinkedList
def remove(index)
if @head.nil?
puts "the storage is empty"
end
if (index==0)
#remove the first element from the list
current=@head
#get the element after the head and make head equal to it
current.next=new_current
@head=new_current
end
if (index>0)
current= get_node(index) #(desired node to be removed)
before_current= get_node(index-1)
after_current=current.next_node
before_current.next_node=after_current
end
end
end
```

```
def get_node(index)
current=@head
index.times do
current=current.next_node
end
return current
end
#this returns a given node based on the index given as argument
```

In my next article, I will discuss how to implement a doubly linked list with Ruby.

for more information on the topic

If you have a question, let me know. If you want to keep up with what I’m doing, follow me on LinkedIn or Twitter