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
import com_github_leetcode.ListNode
/*
* Definition for singly-linked list.
* class ListNode(_x: Int = 0, _next: ListNode = null) {
* var next: ListNode = _next
* var x: Int = _x
* }
*/
object Solution {
def isPalindrome(head: ListNode): Boolean = {
if (head == null || head.next == null) {
return true
}
def reverseList(node: ListNode): ListNode = {
var prev: ListNode = null
var current: ListNode = node
while (current != null) {
val nextNode = current.next
current.next = prev
prev = current
current = nextNode
}
prev
}
def findMiddle(node: ListNode): ListNode = {
var slow = node
var fast = node
while (fast != null && fast.next != null) {
slow = slow.next
fast = fast.next.next
}
slow
}
val middle = findMiddle(head)
var secondHalf = reverseList(middle)
var firstHalf = head
while (secondHalf != null) {
if (firstHalf.x != secondHalf.x) {
return false
}
firstHalf = firstHalf.next
secondHalf = secondHalf.next
}
true
}
}