Reverse a Linked List

In this task, I worked on reversing a singly linked list.
Given the head of a linked list, the goal is to reverse the direction of all the nodes and return the new head.

What I Did
I created a function reverseList that:
Takes the head of a linked list
Reverses the links between nodes
Returns the new head of the reversed list
How I Solved It
Instead of creating a new list , I reversed the list in-place using an iterative approach.

Approach
I used three pointers:

• prev → keeps track of the previous node (initially None)
• curr → points to the current node
• next_node → temporarily stores the next node

Logic Behind It
At each step:

1. Store the next node → next_node = curr.next
2. Reverse the link → curr.next = prev
3. Move prev forward → prev = curr
4. Move curr forward → curr = next_node
5. Repeat this until curr becomes None.

How It Works
The algorithm walks through the list once and flips the direction of each link.
The original head becomes the last node
The last node becomes the new head
By the end, prev will be pointing to the new head of the reversed list.

Time Complexity
O(n) → traverse the list once
Space Complexity
O(1) → no extra space used
Example
Original:
1 → 2 → 3 → 4 → None

Reversed:
4 → 3 → 2 → 1 → None

CODE:

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def reverseList(head):
    prev = None
    curr = head

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

    return prev