Find Middle Element of Linked List in DSA Python - Build from Scratch

Find Middle Element of Linked List

📖 Scenario: Imagine you have a line of people waiting for a bus. You want to find the person standing exactly in the middle of the line.We will represent this line as a linked list, where each person is a node connected to the next.

🎯 Goal: Build a simple linked list with 5 nodes and find the middle node's value using a step-by-step approach.

📋 What You'll Learn

Create a linked list with exactly 5 nodes with values 10, 20, 30, 40, 50 in that order

Create a variable called slow and set it to the head of the linked list

Use a loop to move slow to the middle node by moving two pointers: slow moves one step, fast moves two steps

Print the value of the middle node stored in slow.data

💡 Why This Matters

🌍 Real World

Finding the middle element is useful in many real-world problems like splitting data, balancing tasks, or finding median values.

💼 Career

Understanding linked lists and pointer techniques is important for software engineering roles, especially those involving data structures and algorithms.

1

Create the linked list with 5 nodes

Create a class called Node with an __init__ method that takes data and sets self.data and self.next. Then create 5 nodes with values 10, 20, 30, 40, 50 and link them in order. Finally, create a variable called head that points to the first node.

DSA Python

# Define Node class and create linked list nodes
# Your code here

Hint

Start by defining the Node class with data and next attributes. Then create each node and link them by setting the next attribute.

2

Set up pointers to find the middle

Create two variables called slow and fast and set both to head. These will help us find the middle node.

DSA Python

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

node1 = Node(10)
node2 = Node(20)
node3 = Node(30)
node4 = Node(40)
node5 = Node(50)

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

head = node1

# Set slow and fast pointers to head
# Your code here

Hint

Both slow and fast start at the beginning of the list.

3

Move pointers to find the middle node

Use a while loop that continues as long as fast and fast.next are not None. Inside the loop, move slow one step forward (slow = slow.next) and fast two steps forward (fast = fast.next.next).

DSA Python

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

node1 = Node(10)
node2 = Node(20)
node3 = Node(30)
node4 = Node(40)
node5 = Node(50)

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

head = node1

slow = head
fast = head

# Move slow by one and fast by two steps until fast reaches the end
# Your code here

Hint

Move slow one step and fast two steps in each loop until fast reaches the end.

4

Print the middle node's value

Print the value of the middle node by writing print(slow.data).

DSA Python

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

node1 = Node(10)
node2 = Node(20)
node3 = Node(30)
node4 = Node(40)
node5 = Node(50)

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

head = node1

slow = head
fast = head

while fast is not None and fast.next is not None:
    slow = slow.next
    fast = fast.next.next

# Print the middle node's data
# Your code here

Hint

The middle node's value is stored in slow.data. Print it to see the result.