Reverse a Linked List

Reversing a Singly Linked List

Problem

You’re given the head of a singly linked list. The task is to reverse the list and return the new head.

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Strategy

Reversing a linked list sounded confusing at first because everything points in one direction.

Instead of thinking about the whole list, I focused on one node at a time:

• Keep track of the previous node
• Reverse the current node’s pointer
• Move forward

So the idea is simple — change the direction step by step.

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Code

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

        while current:
            next_node = current.next
            current.next = prev
            prev = current
            current = next_node

        return prev

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Key Lines Explained

• prev = None
  This becomes the new end of the list. Eventually, the original head will point to this.

• current = head
  Start traversing from the beginning.

• next_node = current.next
  Save the next node before changing anything. Without this, the rest of the list would be lost.

• current.next = prev
  This is where the actual reversal happens — the direction of the link is flipped.

• prev = current
  Move prev forward to keep track of the new head.

• current = next_node
  Move ahead in the original list.

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Why This Works

Each step reverses one link. Over time, the entire list gets reversed without creating any new nodes.

It’s just careful pointer manipulation.

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Complexity

• Time: O(n)
• Space: O(1)

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Final Note

This problem feels tricky until you stop thinking about the whole list.

Once you focus on one node at a time, it becomes a simple, repeatable process.