Reverse a Linked List (LeetCode 206) — Explained with Code and Diagrams

Reversing a singly linked list is a classic problem that tests your understanding of pointers and in-place modifications. In this post, I’ll walk you through an iterative solution, explain the pointer logic, and visualize what’s happening under the hood.

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🧠 Problem Statement

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

Example:

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

You must reverse the list in-place, without allocating a new list.

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💡 Core Idea

In a normal singly linked list, each node points to the next:

csharpCopyEdit1 → 2 → 3 → 4 → 5 → null

We need to reverse the links, so it becomes:

csharpCopyEditnull ← 1 ← 2 ← 3 ← 4 ← 5

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🧪 Iterative Solution

✅ Code (Java)

javaCopyEditclass Solution {
    public ListNode reverseList(ListNode head) {
        ListNode pre = null;       // The previous node (starts as null)
        ListNode cur = head;       // The current node being processed
        ListNode temp = null;      // Temporary pointer to store next node

        while (cur != null) {
            temp = cur.next;       // Step 1: Save next node
            cur.next = pre;        // Step 2: Reverse current node's pointer
            pre = cur;             // Step 3: Move pre forward
            cur = temp;            // Step 4: Move cur forward
        }

        return pre;  // pre will be the new head of the reversed list
    }
}

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🔁 Visual Walkthrough

Let’s walk through the reversal of [1 → 2 → 3 → null]:

After the loop, we return pre which now points to the reversed list.

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📊 Time and Space Complexity