Reverse Linked List — DSA Visualizer

Problem

Given the head of a singly linked list, reverse the list, and return the reversed list. Do it in-place with O(1) extra space.

Example 1

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

Output: [5,4,3,2,1]

Example 2

Input: head = [1,2]

Output: [2,1]

Constraints: 0 ≤ n ≤ 5000 | −5000 ≤ Node.val ≤ 5000

Linked List

List reversed — new head =

Pointer State

null

curr

—

nxt

—

Step Logic

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Algorithm

Init prev = null, curr = head

Loop while curr != null

Save nxt = curr.next (don't lose the rest of the list)

Flip pointer: curr.next = prev

Advance: prev = curr, curr = nxt

Return prev — the new head of the reversed list

Time

O(n)

Space

O(1)

Why It Works

At every step, everything behind curr is already a valid reversed sub-list whose new head is prev. By saving nxt before clobbering curr.next, we never lose the forward chain. When curr reaches null, the entire list has been reversed and prev points to the new head.