Category: Programming

The linked list exercises continue! In case you missed them, be sure to check out part 1 and part 2 of this linked list series before moving forward…

Adding an item to the start of a linked list

In this article, the first new feature we’ll implement is the ability to add new items to the start of the list. It’s similar to adding an item to the end of a list, which is why I reviewed it above.

The diagram below shows the first two nodes of a linked list. Recall that a linked list’s head points to its first node and that every node points to the next one:

Here’s how you add a new node to the start of the list. The new node becomes the first node, and the former first node becomes the second node:

Here are the steps:

• Create a new node. In our implementation, newly-created nodes have a next property that points to null, nil, or None by default.
• If the list is empty (i.e. head points to null, nil, or None), adding it to the list is the same thing as adding it to the start of the list. Point head to the new node and you’re done.
• If the list is not empty, you need to do just a little more work.
  • Point the new node’s next property to head. This makes the current first node the one that comes after the new node.
  • Point head at the new node. This makes the new node the first one in the list.

Here’s how I implemented this in Python…

# Python

def add_first(self, value):
    new_node = Node(value)

    # If the list is empty,
    # point `head` to the newly-added node
    # and exit.
    if self.head is None:
        self.head = new_node
        return

    # If the list isn’t empty,
    # make the current first node the second node
    # and make the new node the first node.
    new_node.next = self.head
    self.head = new_node

…and here’s my JavaScript implementation:

# JavaScript

addFirst(value) {
    let newNode = new Node(value)
    
    // If the list is empty,
    // point `head` to the newly-added node
    // and exit.
    if (this.head === null) {
        this.head = newNode
        return
    }
    
    // If the list isn’t empty,
    // make the current first node the second node
    // and make the new node the first node.
    newNode = this.head
    this.head = newNode
}

Inserting a new item at a specified position in the list

So far, we’ve implemented methods to:

• Add a new item to the start of the list, and
• Add a new item to the end of the list.

Let’s do something a little more complicated: adding a new item at a specified position in the list. The position is specified using a number, where 0 denotes the first position in the list.

Here’s how you add a new node at a specified position in the list.

Here are the steps:

• The function has two parameters:
  1. index: The index where a new node should be inserted into the list, with 0 denoting the first item.
  2. value: The value that should be stored in the new node.
• Create a new node, new_node, and put value into it.
• Use a variable, current_index, to keep track of the index of the node we’re currently looking at as we traverse the list. It should initially be set to 0.
• Use another variable, current_node, which will hold a reference to the node we’re currently looking at. It should initially be set to head, which means that we’ll start at the first item in the list.
• Repeat the following as long as current_node is not referencing null, nil, or None:
  • We’re looking for the node in the position immediately before the position where we want to insert the new node. If current_index is one less than index:
    • Point the new node at the current node’s next node.
    • Point the current node at the new node.
    • Return true, meaning that we successfully inserted a new item into the list
  • If we’re here, it means that we haven’t yet reached the node immediately before the insertion point. Move to the next node and increment current_index.
• If we’re here, it means that we’ve gone past the last item in the list. Return false, meaning we didn’t insert a new item into the list.

Here’s how I implemented this in Python…

# Python

def insert_element_at(self, index, value):
    new_node = Node(value)
    current_index = 0
    current_node = self.head

    # Traverse the list, keeping count of 
    # the nodes that you visit,
    # until you’ve reached the specified node.  
    while current_node:
        if current_index == index - 1:
            # We’re at the node before the insertion point!
            # Make the new node point to the next node
            # and the current node point to the new node.
            new_node.next = current_node.next
            current_node.next = new_node
            return True
        
        # We’re not there yet...
        current_node = current_node.next
        current_index += 1
        
    # If you’re here, the given index is larger
    # than the number of elements in the list.
    return False

…and here’s my JavaScript implementation:

// JavaScript

insertElementAt(index, value) {
    let newNode = new Node(value)
    let currentIndex = 0
    let currentNode = this.head
    
    // Traverse the list, keeping count of 
    // the nodes that you visit,
    // until you’ve reached the specified node.     
    while (currentNode) {
        if (currentIndex === index - 1) {
            // We’re at the node before the insertion point!
            // Make the new node point to the next node
            // and the current node point to the new node.
            newNode.next = currentNode.next
            currentNode.next = newNode
            return true
        }
        
        // We’re not there yet...
        currentNode = currentNode.next
        currentIndex++
    }
    
    // If you’re here, the given index is larger
    // than the number of elements in the list.
    return false        
}

Deleting an item from a specified position in the list

Now that we can add a new item at a specified position in the list, let’s implement its opposite: deleting an item from a specified position in the list. Once again, the position is specified using a number, where 0 denotes the first position in the list.

Here’s how you delete a new node from a specified position in the list:

Here are the steps:

• The function has a single parameter, index: the index of the node to be deleted, with 0 denoting the first item in the list.
• Use a variable, current_index, to keep track of the index of the node we’re currently looking at as we traverse the list. It should initially be set to 0.
• Use another variable, current_node, which will hold a reference to the node we’re currently looking at. It should initially be set to head, which means that we’ll start at the first item in the list.
• Repeat the following as long as current_node is not referencing null, nil, or None:
  • We’re looking for the node in the position immediately before the position of the node we want to delete. If current_index is one less than index:
    • Point the current node to the next node’s next node, which removes the next node from the list.
    • Set both the next node’s data and next properties to null, nil, or None. At this point, this node is no longer in the list and is an “island” — no object points to it, and it doesn’t point to any objects. It will eventually be garbage collected.
    • Return true, meaning that we successfully deleted the item from the list
  • If we’re here, it means that we haven’t yet reached the node immediately before the deletion point. Move to the next node and increment current_index.
• If we’re here, it means that we’ve gone past the last item in the list. Return false, meaning we didn’t delete the item from the list.

Here’s how I implemented this in Python…

# Python

def delete_element_at(self, index):
    current_index = 0
    current_node = self.head

    # Traverse the list, keeping count of 
    # the nodes that you visit,
    # until you’ve reached the specified node.  
    while current_node:
        if current_index == index - 1:
            # We’re at the node before the node to be deleted!
            # Make the current node point to the next node’s next node
            # and set the next node’s `data` and `next` properties
            # to `None`.
            delete_node = current_node.next
            current_node.next = delete_node.next
            delete_node.data = None
            delete_node.next = None
            return True
        
        # We’re not there yet...
        current_node = current_node.next
        current_index += 1
        
    # If you’re here, the given index is larger
    # than the number of elements in the list.
    return False

…and here’s my JavaScript implementation:

// JavaScript

deleteElementAt(index) {
    let currentIndex = 0
    let currentNode = this.head
    
    // Traverse the list, keeping count of 
    // the nodes that you visit,
    // until you’ve reached the specified node.  
    while (currentNode) {
        if (currentIndex === index - 1) {
            // We’re at the node before the node to be deleted!
            // Make the current node point to the next node’s next node
            // and set the next node’s `data` and `next` properties
            // to `null`.
            const deleteNode = currentNode.next
            currentNode.next = deleteNode.next
            deleteNode.data = null
            deleteNode.next = null
            return true
        }
        
        // We’re not there yet...
        currentNode = currentNode.next
        currentIndex++
    }
    
    // If you’re here, the given index is larger
    // than the number of elements in the list.
    return false  
}

The classes so far

Let’s take a look at the complete Node and LinkedList classes so far, in both Python and JavaScript.

Python

# Python

class Node:
    
    def __init__(self, data):
        self.data = data
        self.next = None
        
    def __str__(self):
        return f"{self.data}"


class LinkedList:
    
    def __init__(self):
        self.head = None
        
    def __str__(self):
        if self.head is None:
            return('Empty list.')
        
        result = ""
        current_node = self.head
        while current_node:
            result += f'{current_node}\n'
            current_node = current_node.next
            
        return result.strip()
    
    def add_last(self, value):
        new_node = Node(value)

        # If the list is empty,
        # point `head` to the newly-added node
        # and exit.
        if self.head is None:
            self.head = new_node
            return
        
        # If the list isn’t empty,
        # traverse the list by going to each node’s
        # `next` node until there isn’t a `next` node...
        current_node = self.head
        while current_node.next:
            current_node = current_node.next
        
        # If you’re here, `current_node` is
        # the last node in the list.
        # Point `current_node` at
        # the newly-added node.
        current_node.next = new_node
        
    def __len__(self):
        count = 0
        current_node = self.head
        
        # Traverse the list, keeping count of 
        # the nodes that you visit,
        # until you’ve gone past the last node.
        while current_node:
            current_node = current_node.next
            count += 1
            
        return count        
        
    def get_element_at(self, index):
        current_index = 0
        current_node = self.head
        
        # Traverse the list, keeping count of 
        # the nodes that you visit,
        # until you’ve reached the specified node.        
        while current_node:
            if current_index == index:
                # We’re at the node at the given index!
                return current_node.data
            
            # We’re not there yet...
            current_node = current_node.next
            current_index += 1
    
        # If you’re here, the given index is larger
        # than the number of elements in the list.
        return None
    
    def set_element_at(self, index, value):
        current_index = 0
        current_node = self.head
        
        # Traverse the list, keeping count of 
        # the nodes that you visit,
        # until you’ve reached the specified node.  
        while current_node:
            if current_index == index:
                # We’re at the node at the given index!
                current_node.data = value
                return True
            
            # We’re not there yet...
            current_node = current_node.next
            current_index += 1
            
        # If you’re here, the given index is larger
        # than the number of elements in the list.
        return False
        
    def add_first(self, value):
        new_node = Node(value)

        # If the list is empty,
        # point `head` to the newly-added node
        # and exit.
        if self.head is None:
            self.head = new_node
            return
        
        # If the list isn’t empty,
        # make the current first node the second node
        # and make the new node the first node.
        new_node.next = self.head
        self.head = new_node
        
    def insert_element_at(self, index, value):
        new_node = Node(value)
        current_index = 0
        current_node = self.head
        
        # Traverse the list, keeping count of 
        # the nodes that you visit,
        # until you’ve reached the specified node.  
        while current_node:
            if current_index == index - 1:
                # We’re at the node before the insertion point!
                # Make the new node point to the next node
                # and the current node point to the new node.
                new_node.next = current_node.next
                current_node.next = new_node
                return True
            
            # We’re not there yet...
            current_node = current_node.next
            current_index += 1
            
        # If you’re here, the given index is larger
        # than the number of elements in the list.
        return False
        
    def delete_element_at(self, index):
        current_index = 0
        current_node = self.head
        
        # Traverse the list, keeping count of 
        # the nodes that you visit,
        # until you’ve reached the specified node.  
        while current_node:
            if current_index == index - 1:
                # We’re at the node before the node to be deleted!
                # Make the current node point to the next node’s next node
                # and set the next node’s `data` and `next` properties
                # to `None`.
                delete_node = current_node.next
                current_node.next = delete_node.next
                delete_node.data = None
                delete_node.next = None
                return True
            
            # We’re not there yet...
            current_node = current_node.next
            current_index += 1
            
        # If you’re here, the given index is larger
        # than the number of elements in the list.
        return False

JavaScript

// JavaScript

class Node {
    
    constructor(data) {
        this.data = data
        this.next = null
    }
    
    toString() {
        return this.data.toString()
    }
    
}


class LinkedList {
    
    constructor() {
        this.head = null
    }
    
    toString() {
        if (!this.head) {
            return('Empty list.')
        }
        
        let output = ""
        let currentNode = this.head

        while (currentNode) {
            output += `${currentNode.data.toString()}\n`
            currentNode = currentNode.next
        }
        
        return output.trim()
    }
    
    addLast(value) {
        let newNode = new Node(value)
        
        // If the list is empty,
        // point `head` to the newly-added node
        // and exit.
        if (this.head === null) {
            this.head = newNode
            return
        }
        
        // If the list isn’t empty,
        // traverse the list by going to each node’s
        // `next` node until there isn’t a `next` node...
        let currentNode = this.head
        while (currentNode.next) {
            currentNode = currentNode.next
        }
        
        // If you’re here, `currentNode` is
        // the last node in the list.
        // Point `currentNode` at
        // the newly-added node.
        currentNode.next = newNode
    }
    
    getCount() {
        let count = 0
        let currentNode = this.head
        
        // Traverse the list, keeping count of 
        // the nodes that you visit,
        // until you’ve gone past the last node.
        while (currentNode) {
            currentNode = currentNode.next
            count++
        }
        
        return count
    }
    
    getElementAt(index) {
        let currentIndex = 0
        let currentNode = this.head
        
        // Traverse the list, keeping count of 
        // the nodes that you visit,
        // until you’ve reached the specified node.     
        while (currentNode) {
            if (currentIndex === index) {
                // We’re at the node at the given index!
                return currentNode.data
            }
            
            // We’re not there yet...
            currentNode = currentNode.next
            currentIndex++
        }
        
        // If you’re here, the given index is larger
        // than the number of elements in the list.
        return null
    }
    
    setElementAt(index, value) {
        let currentIndex = 0
        let currentNode = this.head
        
        // Traverse the list, keeping count of 
        // the nodes that you visit,
        // until you’ve reached the specified node.     
        while (currentNode) {
            if (currentIndex === index) {
                // We’re at the node at the given index!
                currentNode.data = value
                return true
            }
            
            // We’re not there yet...
            currentNode = currentNode.next
            currentIndex++
        }
        
        // If you’re here, the given index is larger
        // than the number of elements in the list.
        return false
    }
    
    addFirst(value) {
        let newNode = new Node(value)
        
        // If the list is empty,
        // point `head` to the newly-added node
        // and exit.
        if (this.head === null) {
            this.head = newNode
            return
        }
        
        // If the list isn’t empty,
        // make the current first node the second node
        // and make the new node the first node.
        newNode = this.head
        this.head = newNode
    }
    
    insertElementAt(index, value) {
        let newNode = new Node(value)
        let currentIndex = 0
        let currentNode = this.head
        
        // Traverse the list, keeping count of 
        // the nodes that you visit,
        // until you’ve reached the specified node.     
        while (currentNode) {
            if (currentIndex === index - 1) {
                // We’re at the node before the insertion point!
                // Make the new node point to the next node
                // and the current node point to the new node.
                newNode.next = currentNode.next
                currentNode.next = newNode
                return true
            }
            
            // We’re not there yet...
            currentNode = currentNode.next
            currentIndex++
        }
        
        // If you’re here, the given index is larger
        // than the number of elements in the list.
        return false        
    }
    
    deleteElementAt(index) {
        let currentIndex = 0
        let currentNode = this.head
        
        // Traverse the list, keeping count of 
        // the nodes that you visit,
        // until you’ve reached the specified node.  
        while (currentNode) {
            if (currentIndex === index - 1) {
                // We’re at the node before the node to be deleted!
                // Make the current node point to the next node’s next node
                // and set the next node’s `data` and `next` properties
                // to `null`.
                const deleteNode = currentNode.next
                currentNode.next = deleteNode.next
                deleteNode.data = null
                deleteNode.next = null
                return true
            }
            
            // We’re not there yet...
            currentNode = currentNode.next
            currentIndex++
        }
        
        // If you’re here, the given index is larger
        // than the number of elements in the list.
        return false  
    }
    
}

Coming up next

It’s time to reverse that list!

Previously in this series

• How to solve coding interview questions: Linked lists, part 2
• How to solve coding interview questions: Linked lists, part 1
• How to solve coding interview questions: Finding the first NON-recurring character in a string in Swift
• How to solve coding interview questions: Looking deeper into finding the first NON-recurring character in a string
• How to solve coding interview questions: The first NON-recurring character in a string
• Coding interview questions: “First recurring character” in Swift
• How to solve coding interview questions: The first recurring character in a string
• I Has the Dumb (or: How I Embarrassed Myself in My Interview with Google) — from way back in 2013

In case you missed part 1…

Before you continue reading, be sure to check out part 1! It’s where I explain what a linked list is, and provide a couple of basic classes that we’ll build on in this article.

A quick recap

A linked list is a data structure that holds data as a list of items in a specific order. Each item in the list is represented by a node, which “knows” two things:

• Its data
• Where the next node is

Here’s a picture of a node’s structure:

Linked lists are created by connecting nodes together, as shown in the diagram below for a list containing the characters a, b, c, and d in that order:

The diagram above features the following:

• Four nodes, each one for a letter in the list.
• A variable called head, which is a pointer or reference to the first item in the list. It’s the entry point to the list, and it’s a necessary part of the list — you can’t access a linked list if you can’t access its head.
• An optional variable called tail, which is a pointer or reference to the last item in the list. It’s not absolutely necessary, but it’s convenient if you use the list like a queue, where the last element in the list is also the last item that was added to the list.

The JavaScript version of the previous article’s code

In the previous article, I provided code for the Node and LinkedList classes in Python. As promised, here are those classes implemented in JavaScript.

// JavaScript

class Node {
    
    constructor(data) {
        this.data = data
        this.next = null
    }
    
    toString() {
        return this.data.toString()
    }
    
}


class LinkedList {
    
    constructor() {
        this.head = null
    }
    
    toString() {
        let output = ""
        let currentNode = this.head

        while (currentNode) {
            output += `${currentNode.data.toString()}\n`
            currentNode = currentNode.next
        }
        
    toString() {
        if (!this.head) {
            return('Empty list.')
        }
        
        let output = ""
        let currentNode = this.head

        while (currentNode) {
            output += `${currentNode.data.toString()}\n`
            currentNode = currentNode.next
        }
        
        return output.trim()
    }
    
    addLast(data) {
        let newNode = new Node(data)
        
        // If the list is empty,
        // point `head` to the newly-added node
        // and exit.
        if (this.head === null) {
            this.head = newNode
            return
        }
        
        // If the list isn’t empty,
        // traverse the list by going to each node’s
        // `next` node until there isn’t a `next` node...
        let currentNode = this.head
        while (currentNode.next) {
            currentNode = currentNode.next
        }
        
        // If you’re here, `currentNode` is
        // the last node in the list.
        // Point `currentNode` at
        // the newly-added node.
        currentNode.next = newNode
    }

}

…where each item in a list is represented by the node. A node knows only about the data it contains and where the next node in the list is.

The JavaScript version of Node and LinkedList uses the same algorithms as the Python version. The Python version follows the snake_case naming convention, while the JavaScript version follows the camelCase convention.

The JavaScript LinkedList class has the following members:

• head: A property containing a pointer or reference to the first item in the list, or null if the list is empty. This property has the same name in the Python version.
• constructor(): The class constructor, which initializes head to null, meaning that any newly created LinkedList instance is an empty list. In the Python version, the __init__() method has this role.
• toString(): Returns a string representation of the contents of the linked list for debugging purposes. If the list contains at least one item, it returns the contents of the list. If the list is empty, it returns the string Empty list. In the Python version, the __str__() method has this role.
• addLast(): Given a value, it creates a new Node containing that value and adds that Node to the end of the list. In the Python version, the add_last() method has this role.

Adding more functionality to the list

In this article, we’re going to add some much-needed additional functionality to the LinkedList class, namely:

• Reporting how many elements are in the list
• Getting the value of the nth item in the list
• Setting the value of the nth item in the list

I’ll provide both Python and JavaScript implementations.

How many elements are in the list?

To find out how many elements are in a linked list, you have to start at its head and “step through” every item in the list by following each node’s next property, keeping a count of each node you visit. When you reach the last node — the one whose next property is set to None in Python or null in JavaScript — report the number of nodes you counted.

The Python version uses one of its built-in “dunder” (Pythonese for double-underscore) methods, __len__(), to report the number of items in the list. Defining __len__() for LinkedList means that any LinkedList instance will report the number of items it contains when passed as an argument to the built-in len() function. For example, if a LinkedList instance named my_list contains 5 items, len(my_list) returns the value 5.

Here’s the Python version, which you should add to the Python version of LinkedList:

# Python

def __len__(self):
    count = 0
    current_node = self.head
    
    # Traverse the list, keeping count of 
    # the nodes that you visit,
    # until you’ve gone past the last node.
    while current_node:
        current_node = current_node.next
        count += 1
        
    return count    

The JavaScript version is getCount(). If a LinkedList instance named myList contains 5 items, len(myList) returns the value 5.

Here’s the JavaScript version, which you should add to the JavaScript version of LinkedList:

// JavaScript

getCount() {
    let count = 0
    let currentNode = this.head
    
    // Traverse the list, keeping count of 
    // the nodes that you visit,
    // until you’ve gone past the last node.
    while (currentNode) {
        currentNode = currentNode.next
        count++
    }
    
    return count
}

How do I get the data at the nth node in the list?

To get the data at the nth node of the list, you can use an approach that’s similar to the one for counting the list’s elements. You have to start at the head and “step through” every item in the list by following each node’s next property, keeping a count of each node you visit and comparing it against n, where n is the index of the node whose data you want. When the count of nodes you’ve visited is equal to n, report the data contained within the current node.

Here’s the Python version, get_element_at(), which you should add to the Python version of LinkedList:

# Python

def get_element_at(self, index):
    current_index = 0
    current_node = self.head

    # Traverse the list, keeping count of 
    # the nodes that you visit,
    # until you’ve reached the specified node.        
    while current_node:
        if current_index == index:
            # We’re at the node at the given index!
            return current_node.data
        
        # We’re not there yet...
        current_node = current_node.next
        current_index += 1

    # If you’re here, the given index is larger
    # than the number of elements in the list.
    return None

Here’s the JavaScript version, getElementAt(), which you should add to the JavaScript version of LinkedList:

// JavaScript

getElementAt(index) {
    let currentIndex = 0
    let currentNode = this.head
    
    // Traverse the list, keeping count of 
    // the nodes that you visit,
    // until you’ve reached the specified node.     
    while (currentNode) {
        if (currentIndex === index) {
            // We’re at the node at the given index!
            return currentNode.data
        }
        
        // We’re not there yet...
        currentNode = currentNode.next
        currentIndex++
    }
    
    // If you’re here, the given index is larger
    // than the number of elements in the list.
    return null
}

How do I set the value of the nth node in the list?

To set the data at the nth node of the list, you can use an approach that’s similar to the one for getting the data at the nth node. You have to start at the head and “step through” every item in the list by following each node’s next property, keeping a count of each node you visit and comparing it against n, where n is the index of the node whose data you want. When the count of nodes you’ve visited is equal to n, set the current node’s data property to the new value.

Here’s the Python version, set_element_at(), which you should add to the Python version of LinkedList:

# Python

def set_element_at(self, index, value):
    current_index = 0
    current_node = self.head

    # Traverse the list, keeping count of 
    # the nodes that you visit,
    # until you’ve reached the specified node.  
    while current_node:
        if current_index == index:
            # We’re at the node at the given index!
            current_node.data = value
            return True
        
        # We’re not there yet...
        current_node = current_node.next
        current_index += 1
        
    # If you’re here, the given index is larger
    # than the number of elements in the list.
    return False

Here’s the JavaScript version, setElementAt(), which you should add to the JavaScript version of LinkedList:

// JavaScript

setElementAt(index, value) {
    let currentIndex = 0
    let currentNode = this.head
    
    // Traverse the list, keeping count of 
    // the nodes that you visit,
    // until you’ve reached the specified node.     
    while (currentNode) {
        if (currentIndex === index) {
            // We’re at the node at the given index!
            currentNode.data = value
            return true
        }
        
        // We’re not there yet...
        currentNode = currentNode.next
        currentIndex++
    }
    
    // If you’re here, the given index is larger
    // than the number of elements in the list.
    return false
}

The classes so far

Let’s take a look at the complete Node and LinkedList classes so far, in both Python and JavaScript.

Python

# Python

class Node:
    
    def __init__(self, data):
        self.data = data
        self.next = None
        
    def __str__(self):
        return f"{self.data}"


class LinkedList:
    
    def __init__(self):
        self.head = None
        
    def __str__(self):
        if self.head is None:
            return('Empty list.')
        
        result = ""
        current_node = self.head
        while current_node:
            result += f'{current_node}\n'
            current_node = current_node.next
            
        return result.strip()
    
    def add_last(self, value):
        new_node = Node(value)

        # If the list is empty,
        # point `head` to the newly-added node
        # and exit.
        if self.head is None:
            self.head = new_node
            return
        
        # If the list isn’t empty,
        # traverse the list by going to each node’s
        # `next` node until there isn’t a `next` node...
        current_node = self.head
        while current_node.next:
            current_node = current_node.next
        
        # If you’re here, `current_node` is
        # the last node in the list.
        # Point `current_node` at
        # the newly-added node.
        current_node.next = new_node
        
    def __len__(self):
        count = 0
        current_node = self.head
        
        # Traverse the list, keeping count of 
        # the nodes that you visit,
        # until you’ve gone past the last node.
        while current_node:
            current_node = current_node.next
            count += 1
            
        return count        
        
    def get_element_at(self, index):
        current_index = 0
        current_node = self.head
        
        # Traverse the list, keeping count of 
        # the nodes that you visit,
        # until you’ve reached the specified node.        
        while current_node:
            if current_index == index:
                # We’re at the node at the given index!
                return current_node.data
            
            # We’re not there yet...
            current_node = current_node.next
            current_index += 1
    
        # If you’re here, the given index is larger
        # than the number of elements in the list.
        return None
    
    def set_element_at(self, index, value):
        current_index = 0
        current_node = self.head
        
        # Traverse the list, keeping count of 
        # the nodes that you visit,
        # until you’ve reached the specified node.  
        while current_node:
            if current_index == index:
                # We’re at the node at the given index!
                current_node.data = value
                return True
            
            # We’re not there yet...
            current_node = current_node.next
            current_index += 1
            
        # If you’re here, the given index is larger
        # than the number of elements in the list.
        return False

JavaScript

// JavaScript

class Node {
    
    constructor(data) {
        this.data = data
        this.next = null
    }
    
    toString() {
        return this.data.toString()
    }
    
}


class LinkedList {
    
    constructor() {
        this.head = null
    }
    
    toString() {
        if (!this.head) {
            return('Empty list.')
        }
        
        let output = ""
        let currentNode = this.head

        while (currentNode) {
            output += `${currentNode.data.toString()}\n`
            currentNode = currentNode.next
        }
        
        return output.trim()
    }
    
    addLast(value) {
        let newNode = new Node(value)
        
        // If the list is empty,
        // point `head` to the newly-added node
        // and exit.
        if (this.head === null) {
            this.head = newNode
            return
        }
        
        // If the list isn’t empty,
        // traverse the list by going to each node’s
        // `next` node until there isn’t a `next` node...
        let currentNode = this.head
        while (currentNode.next) {
            currentNode = currentNode.next
        }
        
        // If you’re here, `currentNode` is
        // the last node in the list.
        // Point `currentNode` at
        // the newly-added node.
        currentNode.next = newNode
    }
    
    getCount() {
        let count = 0
        let currentNode = this.head
        
        // Traverse the list, keeping count of 
        // the nodes that you visit,
        // until you’ve gone past the last node.
        while (currentNode) {
            currentNode = currentNode.next
            count++
        }
        
        return count
    }
    
    getElementAt(index) {
        let currentIndex = 0
        let currentNode = this.head
        
        // Traverse the list, keeping count of 
        // the nodes that you visit,
        // until you’ve reached the specified node.     
        while (currentNode) {
            if (currentIndex === index) {
                // We’re at the node at the given index!
                return currentNode.data
            }
            
            // We’re not there yet...
            currentNode = currentNode.next
            currentIndex++
        }
        
        // If you’re here, the given index is larger
        // than the number of elements in the list.
        return null
    }
    
    setElementAt(index, value) {
        let currentIndex = 0
        let currentNode = this.head
        
        // Traverse the list, keeping count of 
        // the nodes that you visit,
        // until you’ve reached the specified node.     
        while (currentNode) {
            if (currentIndex === index) {
                // We’re at the node at the given index!
                currentNode.data = value
                return true
            }
            
            // We’re not there yet...
            currentNode = currentNode.next
            currentIndex++
        }
        
        // If you’re here, the given index is larger
        // than the number of elements in the list.
        return false
    }

}

Wait…this “linked list” thing is beginning to look like a Python or JavaScript array, but not as fully-featured. What’s going on here?

The truth about linked lists

Here’s the truth: there’s a better-than-even chance that you’ll never use them…directly.

If you program in most languages from the 1990s or later (for example, Python was first released in 1991, PHP’s from 1994, Java and JavaScript came out in 1995), you’re working with arrays — or in Python’s case, lists, which are almost the same thing — that you can you can add elements to, remove elements from, perform searches on, and do all sorts of things with.

When you use one of these data types in a modern language…

• Python lists or a JavaScript arrays
• Dictionaries, also known as objects in JavaScript, hashes in Ruby, or associative arrays
• Stacks and queues
• React components

…there’s a linked list behind the scenes, moving data around by traversing, adding, and deleting nodes. There’s a pretty good chance that you’re indirectly using linked lists in your day-to-day programming; you’re just insulated from the details by layers of abstractions.

So why do they still pose linked list challenges in coding interviews?

History plays a major factor.

In older languages like FORTRAN (1957), Pascal (1970), and C (1972), arrays weren’t dynamic, but fixed in size. If you declared an array with 20 elements, it remained a 20-element array. You couldn’t add or remove elements at run-time.

But Pascal and C supported pointers from the beginning, and pointers make it possible to create dynamic data structures — the kind where you can add or delete elements at run-time. In those early pre-web days, when programming languages’ standard libraries weren’t as rich and complete as today’s, and you often had to “roll your own” dynamic data structures, such as trees, queues, stacks, and…linked lists.

If you majored in computer science in the ’70s, ’80s, and even the ’90s, you were expected to know how to build your own dynamic data structures from scratch, and they most definitely appeared on the exam. These computer science majors ended up in charge at tech companies, and they expected the people they hired to know this stuff as well. Over time, it became tradition, and to this day, you’ll still get the occasional linked list question in a coding interview.

What’s the point in learning about linked lists when we’re mostly insulated from having to work with them directly?

For starters, there’s just plain old pragmatism. As long as they’re still asking linked list questions in coding interviews, it’s a good idea to get comfortable with them and be prepared to answer all manner of questions about algorithms and data structures.

Secondly, it’s a good way to get better at the kind of problem solving that we need in day-to-day programming. Working with linked lists requires us to think about pointers/references, space and time efficiency, tradeoffs, and even writing readable, maintainable code. And of course, there’s always a chance that you might have to implement a linked list yourself, whether because you’re working on an IoT or embedded programming project with a low-level language like C or implementing something lower-level such as the reconciler in React.

Coming up next

• Adding a new element to the head of a linked list
• Adding a new element at a specific place in the middle of a linked list
• Deleting a specified element of a linked list

Previously in this series

• How to solve coding interview questions: Linked lists, part 1
• How to solve coding interview questions: Finding the first NON-recurring character in a string in Swift
• How to solve coding interview questions: Looking deeper into finding the first NON-recurring character in a string
• How to solve coding interview questions: The first NON-recurring character in a string
• Coding interview questions: “First recurring character” in Swift
• How to solve coding interview questions: The first recurring character in a string
• I Has the Dumb (or: How I Embarrassed Myself in My Interview with Google) — from way back in 2013