Leetcode234 palindrome linked list

First check the title , The so-called palindrome , It's the same for forward traversal and backward traversal , It's easy to think , If I find the intermediate node , The number of each node from the middle to the back and from the middle to the front should be the same . This problem can be said to be an upgraded version of the reverse linked list . We need to use the idea of a speed pointer , If I move the pointer forward two digits at a time , The slow pointer advances one bit at a time , Fast pointer to the end , The slow pointer must be in the middle .

Pictured The length is odd and even

class Solution {
   public boolean isPalindrome(ListNode head) {
        if(head == null || head.next == null) {
            return true;
        }
        ListNode slow = head, fast = head;
        ListNode pre = head, prepre = null;
        while(fast != null && fast.next != null) {
            pre = slow;
            slow = slow.next;
            fast = fast.next.next;
            pre.next = prepre;
            prepre = pre;
        }
        if(fast != null) {
            slow = slow.next;
        }
        while(pre != null && slow != null) {
            if(pre.val != slow.val) {
                return false;
            }
            pre = pre.next;
            slow = slow.next;
        }
        return true;
    }

}