Reversing a Linked List Using Iterative Method in Python

Problem Explanation

You are given the head of a singly linked list.
Your task is to reverse the linked list and return the new head.

A linked list looks like:
1 → 2 → 3 → 4 → 5

After reversing:
5 → 4 → 3 → 2 → 1

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Method Used: Iterative Approach

Idea

We reverse the direction of each node’s pointer one by one.

For each node:

• Store next node
• Reverse the link
• Move forward

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Why This Method?

• Time complexity: O(n)
• Space complexity: O(1)
• Simple and efficient
• No extra memory used

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Python Code with Explanation

class Solution:
    def reverseList(self, head):

Defines the function. head is the starting node.

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        prev = None

prev will store the previous node (initially None).

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        curr = head

Start from the head of the list.

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        while curr:

Loop until we reach the end of the list.

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            next_node = curr.next

Store the next node before breaking the link.

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            curr.next = prev

Reverse the link (point current node to previous).

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            prev = curr

Move prev forward.

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            curr = next_node

Move curr forward.

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        return prev

At the end, prev becomes the new head.

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Complete Code

class Solution:
    def reverseList(self, head):
        prev = None
        curr = head

        while curr:
            next_node = curr.next
            curr.next = prev
            prev = curr
            curr = next_node

        return prev

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Step-by-Step Example

Input:

1 → 2 → 3 → 4 → 5

Steps:

• Reverse links one by one
• Final reversed list

Output:

5 → 4 → 3 → 2 → 1

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Key Takeaway

Reversing a linked list is about changing pointer directions, not values. The iterative approach is the most efficient and commonly used method.

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