Reverse Linked List in Python — Iterative (Optimal) & More

Reverse Linked List (#206 — Easy): Given the head of a singly linked list, reverse the list, and return the reversed list.

Problem Statement

Given the head of a singly linked list, reverse the list, and return the reversed list.

Example:

Input: head = [1,2,3,4,5]
Output: [5,4,3,2,1]

Approach: Iterative (Optimal)

Time: O(n) | Space: O(1)

Three pointers dance: save next, flip current’s pointer backward, advance both. The head becomes the tail and prev ends up as the new head.

def reverse_list(head):
    prev = None
    curr = head
    while curr:
        nxt = curr.next
        curr.next = prev
        prev = curr
        curr = nxt
    return prev

Approach: Recursive

Time: O(n) | Space: O(n) (call stack)

Recursively reverse the rest of the list, then fix the current node. Elegant but uses stack space proportional to list length.

def reverse_list_recursive(head):
    if not head or not head.next:
        return head
    new_head = reverse_list_recursive(head.next)
    head.next.next = head
    head.next = None
    return new_head

Key Takeaways

• Start with the brute-force approach to understand the problem
• Iterative (Optimal) gives the optimal O(n) solution
• Practice this pattern — it appears frequently in coding interviews

Related Problems

• Reverse Linked List in Dart
• Reverse Linked List in Go
• Reverse Linked List in Swift
• Linked List Cycle in Python

FAQ

What is the best approach to solve Reverse Linked List in Python?

The recommended approach is Iterative (Optimal), which runs in O(n) time with O(1) space. The full Python implementation is shown above.

What is the time complexity of Reverse Linked List in Python?

Using Iterative (Optimal), the time complexity is O(n) and the space complexity is O(1).

Happy coding!