Palindrome Linked List — DSA Visualizer

Problem

Given the head of a singly linked list, return true if it is a palindrome, or false otherwise.

Example 1

Input: head = [1,2,2,1]

Output: true

Example 2

Input: head = [1,2]

Output: false

Constraints: 1 ≤ n ≤ 10⁵ | 0 ≤ Node.val ≤ 9

Phase 1 — Find Middle

State Variables

Phase

—

left.val

—

right.val

—

Match?

—

Step Logic

Press ▶ Play or Next Step to begin the animation.

Ready

0 / 0

Select an example above and press Play.

Algorithm

Find Middle: slow advances 1 step, fast advances 2. When fast hits end, slow is at the midpoint.

Reverse 2nd half: Standard in-place reversal from slow. prev ends up as the new right-half head.

Compare halves: left from head, right from prev. Any mismatch → return false.

All pairs matched → return true (it is a palindrome).

Time

O(n)

Space

O(1)

Why This Works

A palindrome reads the same forwards and backwards. By reversing the second half in-place we avoid using a stack (O(n) space). The slow/fast pointer trick finds the midpoint in a single O(n) pass. The final comparison is also O(n). Total: O(n) time, O(1) space.