Get Length of Linked List in DSA Python

Mental Model

Count each node one by one until you reach the end of the list.

Analogy: Like counting people standing in a line by moving from the first person to the last one.

head -> 1 -> 2 -> 3 -> null

Dry Run Walkthrough

Input: list: 1 -> 2 -> 3 -> null

Goal: Find how many nodes are in the linked list

Step 1: start at head node with value 1

head -> [curr->1] -> 2 -> 3 -> null

Why: We begin counting from the first node

Step 2: count node 1 and move curr to next node

head -> 1 -> [curr->2] -> 3 -> null

Why: Counted first node, move to second to continue counting

Step 3: count node 2 and move curr to next node

head -> 1 -> 2 -> [curr->3] -> null

Why: Counted second node, move to third node

Step 4: count node 3 and move curr to null (end)

head -> 1 -> 2 -> 3 -> [curr->null]

Why: Counted last node, reached end of list

Result:

head -> 1 -> 2 -> 3 -> null
Length = 3

Annotated Code

DSA Python

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

class LinkedList:
    def __init__(self):
        self.head = None

    def append(self, val):
        new_node = Node(val)
        if not self.head:
            self.head = new_node
            return
        curr = self.head
        while curr.next:
            curr = curr.next
        curr.next = new_node

    def get_length(self):
        count = 0
        curr = self.head
        while curr:
            count += 1  # count current node
            curr = curr.next  # move to next node
        return count

# Driver code
ll = LinkedList()
ll.append(1)
ll.append(2)
ll.append(3)
length = ll.get_length()
print(f"Length = {length}")

while curr:

loop through each node until end of list

count += 1 # count current node

increment count for each node visited

curr = curr.next # move to next node

advance pointer to next node to continue counting

Output

Length = 3

Complexity Analysis

Time: O(n) because we visit each node once to count

Space: O(1) because we only use a few variables for counting

vs Alternative: Alternative would be storing length in a variable updated on insert/delete, which uses extra space but gives O(1) length retrieval

Edge Cases

empty list

returns length 0 because head is None and loop never runs

DSA Python

while curr:

single node list

returns length 1 by counting the single node

DSA Python

while curr:

When to Use This Pattern

When you need to know how many elements are in a linked list, use a simple traversal counting each node until the end.

Common Mistakes

Mistake: Forgetting to move the current pointer to the next node inside the loop
Fix: Add 'curr = curr.next' inside the while loop to advance the pointer

Mistake: Initializing count to 1 instead of 0, causing off-by-one error
Fix: Initialize count to 0 before starting the loop

Summary

Counts the number of nodes in a linked list by visiting each node once.

Use when you need to find the size of a linked list without stored length.

The key is to start at the head and move through each node until you reach the end.