DSA Linked Lists Operations

Linked List Operations

Basic things we can do with linked lists are:

1. Traversal
2. Remove a node
3. Insert a node
4. Sort

For simplicity, singly linked lists will be used to explain these operations below.

Traversal of a Linked List

Traversing a linked list means to go through the linked list by following the links from one node to the next.

Traversal of linked lists is typically done to search for a specific node, and read or modify the node's content, remove the node, or insert a node right before or after that node.

To traverse a singly linked list, we start with the first node in the list, the head node, and follow that node's next link, and the next node's next link and so on, until the next address is null, like in the animation below:

The code below prints out the node values as it traverses along the linked list, in the same way as the animation above.

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

def traverseAndPrint(head):
    currentNode = head
    while currentNode:
        print(currentNode.data, end=" -> ")
        currentNode = currentNode.next
    print("null")

node1 = Node(7)
node2 = Node(11)
node3 = Node(3)
node4 = Node(2)
node5 = Node(9)

node1.next = node2
node2.next = node3
node3.next = node4
node4.next = node5

traverseAndPrint(node1)

Find The Lowest Value in a Linked List

Let's find the lowest value in a singly linked list by traversing it and checking each value.

Finding the lowest value in a linked list is very similar to how we found the lowest value in an array, except that we need to follow the next link to get to the next node.

This is how finding the lowest value in a linked list works in principle:

To find the lowest value we need to traverse the list like in the previous code. But in addition to traversing the list, we must also update the current lowest value when we find a node with a lower value.

In the code below, the algorithm to find the lowest value is moved into a function called findLowestValue.

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

def findLowestValue(head):
    minValue = head.data
    currentNode = head.next
    while currentNode:
        if currentNode.data < minValue:
            minValue = currentNode.data
        currentNode = currentNode.next
    return minValue

node1 = Node(7)
node2 = Node(11)
node3 = Node(3)
node4 = Node(2)
node5 = Node(9)

node1.next = node2
node2.next = node3
node3.next = node4
node4.next = node5

print("The lowest value in the linked list is:", findLowestValue(node1))

Delete a Node in a Linked List

In this case we have the link (or pointer or address) to a node that we want to delete.

It is important to connect the nodes on each side of the node before deleting it, so that the linked list is not broken.

So before deleting the node, we need to get the next pointer from the previous node, and connect the previous node to the new next node before deleting the node in between.

In a singly linked list, like we have here, to get the next pointer from the previous node we actually need traverse the list from the start, because there is no way to go backwards from the node we want to delete.

The simulation below shows the node we want to delete, and how the list must be traversed first to connect the list properly before deleting the node without breaking the linked list.

Also, it is a good idea to first connect next pointer to the node after the node we want to delete, before we delete it. This is to avoid a 'dangling' pointer, a pointer that points to nothing, even if it is just for a brief moment.

In the code below, the algorithm to delete a node is moved into a function called deleteSpecificNode.

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

def traverseAndPrint(head):
    currentNode = head
    while currentNode:
        print(currentNode.data, end=" -> ")
        currentNode = currentNode.next
    print("null")

def deleteSpecificNode(head, nodeToDelete):

    if head == nodeToDelete:
        return head.next

    currentNode = head
    while currentNode.next and currentNode.next != nodeToDelete:
        currentNode = currentNode.next

    if currentNode.next is None:
        return head

    currentNode.next = currentNode.next.next

    return head

node1 = Node(7)
node2 = Node(11)
node3 = Node(3)
node4 = Node(2)
node5 = Node(9)

node1.next = node2
node2.next = node3
node3.next = node4
node4.next = node5

print("Before deletion:")
traverseAndPrint(node1)

# Delete node4
node1 = deleteSpecificNode(node1, node4)

print("\nAfter deletion:")
traverseAndPrint(node1)

In the deleteSpecificNode function above, the return value is the new head of the linked list. So for example, if the node to be deleted is the first node, the new head returned will be the next node.

Insert a Node in a Linked List

Inserting a node into a linked list is very similar to deleting a node, because in both cases we need to take care of the next pointers to make sure we do not break the linked list.

To insert a node in a linked list we first need to create the node, and then at the position where we insert it, we need to adjust the pointers so that the previous node points to the new node, and the new node points to the correct next node.

The simulation below shows how the links are adjusted when inserting a new node.

1. New node is created
2. Node 1 is linked to new node
3. New node is linked to next node

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

def traverseAndPrint(head):
    currentNode = head
    while currentNode:
        print(currentNode.data, end=" -> ")
        currentNode = currentNode.next
    print("null")

def insertNodeAtPosition(head, newNode, position):
    if position == 1:
        newNode.next = head
        return newNode
    
    currentNode = head
    for _ in range(position - 2):
        if currentNode is None:
            break
        currentNode = currentNode.next

    newNode.next = currentNode.next
    currentNode.next = newNode
    return head

node1 = Node(7)
node2 = Node(3)
node3 = Node(2)
node4 = Node(9)

node1.next = node2
node2.next = node3
node3.next = node4

print("Original list:")
traverseAndPrint(node1)

# Insert a new node with value 97 at position 2
newNode = Node(97)
node1 = insertNodeAtPosition(node1, newNode, 2)

print("\nAfter insertion:")
traverseAndPrint(node1)

In the insertNodeAtPosition function above, the return value is the new head of the linked list. So for example, if the node is inserted at the start of the linked list, the new head returned will be the new node.

Other Linked Lists Operations

We have only covered three basic linked list operations above: traversal (or search), node deletion, and node insertion.

There are a lot of other operations that could be done with linked lists, like sorting for example.

Previously in the tutorial we have covered many sorting algorithms, and we could do many of these sorting algorithms on linked lists as well. Let's take selection sort for example. In selection sort we find the lowest value, remove it, and insert it at the beginning. We could do the same with a linked list as well, right? We have just seen how to search through a linked list, how to remove a node, and how to insert a node.

Linked Lists vs Arrays

These are some key linked list properties, compared to arrays:

• Linked lists are not allocated to a fixed size in memory like arrays are, so linked lists do not require to move the whole list into a larger memory space when the fixed memory space fills up, like arrays must.
• Linked list nodes are not laid out one right after the other in memory (contiguously), so linked list nodes do not have to be shifted up or down in memory when nodes are inserted or deleted.
• Linked list nodes require more memory to store one or more links to other nodes. Array elements do not require that much memory, because array elements do not contain links to other elements.
• Linked list operations are usually harder to program and require more lines than similar array operations, because programming languages have better built in support for arrays.
• We must traverse a linked list to find a node at a specific position, but with arrays we can access an element directly by writing myArray[5].

Time Complexity of Linked Lists Operations

Here we discuss time complexity of linked list operations, and compare these with the time complexity of the array algorithms that we have discussed previously in this tutorial.

Remember that time complexity just says something about the approximate number of operations needed by the algorithm based on a large set of data \(n\), and does not tell us the exact time a specific implementation of an algorithm takes.

This means that even though linear search is said to have the same time complexity for arrays as for linked list: \(O(n)\), it does not mean they take the same amount of time. The exact time it takes for an algorithm to run depends on programming language, computer hardware, differences in time needed for operations on arrays vs linked lists, and many other things as well.

Linear search for linked lists works the same as for arrays. A list of unsorted values are traversed from the head node until the node with the specific value is found. Time complexity is \(O(n)\).

Binary search is not possible for linked lists because the algorithm is based on jumping directly to different array elements, and that is not possible with linked lists.

Sorting algorithms have the same time complexities as for arrays, and these are explained earlier in this tutorial. But remember, sorting algorithms that are based on directly accessing an array element based on an index, do not work on linked lists.

def traverseAndPrint(head):
    currentNode = 
    while currentNode:
        print(currentNode.data, end=" -> ")
        currentNode = currentNode.
    print("null")