LeetCode-in-All

234. Palindrome Linked List

Easy

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:

• The number of nodes in the list is in the range [1, 105].
• 0 <= Node.val <= 9

Follow up: Could you do it in O(n) time and O(1) space?

Solution

; Definition for singly-linked list:
#|

; val : integer?
; next : (or/c list-node? #f)
(struct list-node
  (val next) #:mutable #:transparent)

; constructor
(define (make-list-node [val 0])
  (list-node val #f))

|#

(define (linked-list->numbers ll)
  (if (not (list-node? ll))
      '()
      (append (list (list-node-val ll))
              (linked-list->numbers (list-node-next ll)))))


(define (is-palindrome head)
  (let ([numls (linked-list->numbers head)])
    (equal? numls (reverse numls))))

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