A list is a collection of elements. A linked List is a list in which each element in the list contains both and a to one or both neighboring items. L 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 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 list has nodes with pointers which always point to the next element in the list. This discussion  focus on would be discussed in my subsequent article. data pointer inked List doubly-linked singly-linked singly-linked list. Doubly-linked list Let us consider the list: 10 →20 →30 We can say points to points to . The element of a list is referred to as a Node. So in this case 10 →20 →30, we would say points to and points to Each node in a singly linked list has a . attribute to indicate the next node after it (the current node). 10 20, 20 30 Node10 Node20, Node20 Node30. next "NOTE: each node has .value for the current value and .next for next node(element)in the list. The first node is the head , in this case, head.value =10 and head.next =20" Why use linked List? 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 collections of indexed by ordered items contiguous integers . How to Add or remove a node(element) from a singly-linked list First, we would create a node to store the data. Nodes of a singly-linked list have two attributes: and value( value of current node ) next_node( node after current ) , @value = value
      @next_node = next_node class Node attr_accessor :value :next_node def initialize (value, next_node = ) nil 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 after 10 the  list below 15 " " 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 Add at the end of the list @head = @tail = this_node = Node.new(number) @head. ?
         @head = this_node current = @head current.next_node. ?
        current = current.next_node current.next_node = this_node #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 nil nil end def add (number) #create a new node if nil return end #until current.nil means until we reach the last node until nil end #point current(last node) to our new 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 ), we will loop up to the position we want to insert at. current.next.nil? @head. ? this_nod=Node.new(item)
        @head=this_nod index== this_nod=Node.new(item,@head)
        @head=this_nod index > ind=index- current=@head
      before_current=@head ind.times before_current=current.next_node index.times current=current.next_node this_nod=Node.new(item) 
      after_current=before_current.next_node before_current.next_node=this_nod this_nod.next_node=after_current #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 nil #if list is empty, the head is the new node end if 0 # if index is 0, we insert in the first position end if 0 #insert at desired position if index is greater than 0 1 #loop to the desired position before where you wish to insert do end #loop to the desired position where you wish to insert do end #create a new node you wish to insert #point node before current to new node #point new node to the old current node end end end Remove at a given position let us consider deleting or removing an element from the a list. Assuming we want to delete(remove) 20 from our list " Note: Make the node with value 10, point to the node with value 30. Afterwards, we make the node with value 20 point to itself(OR NIL). Once the head, in this case, points to the node with value 30, 20 is said to have been removed since it would no longer be accessible from the list." Now let us try to remove an element from the list. @head. ?
        puts (index== ) current=@head current. =new_current 
        @head=new_current (index> )
        current= get_node(index) before_current= get_node(index- ) 
        after_current=current.next_node
        before_current.next_node=after_current #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 nil "the storage is empty" end if 0 #remove the first element from the list #get the element after the head and make head equal to it next end if 0 #(desired node to be removed) 1 end end end " Note : the get_node method is a customized private method that get the node at a given position.This is to prevent repetition." current=@head
    index.times current=current.next_node current def get_node (index) do end return 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. Conclusion 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 or LinkedIn Twitter

Twitter

Read My Stories

Too Long; Didn't Read

In your car, at home, or at work — Bosch technology shapes many areas of life.

How to Implement Singly Linked List with Ruby

Too Long; Didn't Read

Company Mentioned

Addo Boakye Forison

About Author

TOPICS

THIS ARTICLE WAS FEATURED IN...

How to Implement Singly Linked List with Ruby

Too Long; Didn't Read

Company Mentioned

Addo Boakye Forison

About Author

TOPICS

THIS ARTICLE WAS FEATURED IN...

RELATED STORIES