Reverse Linked List: Pointer reversal Pattern

Learning Outcome

After this lesson, you should be able to reverse a linked list in-place while carefully preserving the next node before changing pointers.

Problem Statement

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

Brute Force Approach

Copy all node values into an array, then rebuild a new linked list in reverse order.

This is easy to visualize, but it uses O(n) extra space and does not practice pointer manipulation.

Optimized Approach

Use three pointers: prev, curr, and nextNode. Save curr.next, point curr.next to prev, then move both pointers forward.

The key safety rule is to save the next node before changing curr.next.

Exact Pseudocode

prev = null
curr = head
while curr is not null:
  nextNode = curr.next
  curr.next = prev
  prev = curr
  curr = nextNode
return prev
  

Reference 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
  

Sample Dry Run

Complexity

Edge Cases

• Empty list returns null.
• Single-node list returns the same node.
• Saving nextNode is required before changing curr.next.

Interview Checklist

• Save the next node before rewiring.
• Move prev and curr in the right order.
• Return prev, not the old head.

FAQs