What is the brute force approach and complexity for Palindrome Linked List?

Copy values into a vector, compare from both ends with time complexity O(N) and space complexity O(N).

What is the optimal approach and complexity for Palindrome Linked List?

Slow/fast to the middle, reverse the back half in place, compare, reverse it back with optimal time complexity O(N) and space complexity O(1).

How do you identify the pattern for Palindrome Linked List?

Look for key signals in the prompt: "palindrome" indicates Compare the front with the back → split the list into two halves at the middle.; "singly linked list" indicates Can't walk backwards → reverse the back half so it can be walked forwards from the tail.; "O(1) extra space" indicates No copying into a vector or stack → find the middle with slow/fast and relink in place..

Palindrome Linked List

Easy·LeetCode #234 Pro OnlyBasic Reinforcement

LeetCode

Problem Statement

You are given the head of a singly linked list. Return true if the values read the same forwards and backwards (a palindrome), and false otherwise.

Follow-up: can you do it in O(N) time and O(1) extra space?

Input

head, the first node of a singly linked list.

Output

true if the list's values form a palindrome, otherwise false.

Example 1

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

Output: true

Read backwards it is still 1, 2, 2, 1.

Example 2

Input: head = [1, 2]

Output: false

Backwards it is 2, 1, which is different.

Constraints

• The list has between 1 and 10^5 nodes.
• 0 ≤ Node.val ≤ 9
• Follow-up: O(N) time and O(1) extra space.

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“palindrome”

→ Compare the front with the back → split the list into two halves at the middle.

“singly linked list”

→ Can't walk backwards → reverse the back half so it can be walked forwards from the tail.

“O(1) extra space”

→ No copying into a vector or stack → find the middle with slow/fast and relink in place.

Approach Complexity Ladder

Brute Force — copy values to an arrayO(N)O(N)

Copy values into a vector, compare from both ends

Optimal — find the middle, reverse the second half, compare, restoreO(N)O(1)

Slow/fast to the middle, reverse the back half in place, compare, reverse it back

Brute Force Intuition

The trouble with a linked list is that we can't walk it backwards. An array doesn't have that problem. Copy all values into a vector, then use two pointers: one at the start, one at the end, moving inward and comparing.

Recognition rule for this sub-pattern:

When fast reaches the end, slow is in the middle. Use this to split a list in half.

Remember this · Linked List — Fast & Slow

Trigger

Cycle detection, middle of list, "n-th from end", two lists meeting, repeating sequences.

Think

If one pointer moves faster or starts earlier, where will they meet?

Key question

What speed or gap makes the meeting point tell me the answer?

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