LeetCode-in-All
234. Palindrome Linked List
Easy
Given the head of a singly linked list, return true if it is a palindrome.
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.
* class ListNode {
* val: number
* next: ListNode | null
* constructor(val?: number, next?: ListNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
* }
*/
function isPalindrome(head: ListNode | null): boolean {
if (head === null || head.next === null) {
// Empty list or single element is considered a palindrome.
return true
}
let len = 0
let right = head
// Calculate the length of the list.
while (right !== null) {
right = right.next
len++
}
// Find the middle of the list.
let middle = Math.floor(len / 2)
// Reset the right pointer to the head.
right = head
// Move the right pointer to the middle of the list.
for (let i = 0; i < middle; i++) {
right = right.next
}
// Reverse the right half of the list.
let prev = null
while (right !== null) {
const next = right.next
right.next = prev
prev = right
right = next
}
// Compare the left and right halves.
for (let i = 0; i < middle; i++) {
if (prev !== null && head.val === prev.val) {
head = head.next
prev = prev.next
} else {
return false
}
}
return true
}
export { isPalindrome }