palindrome linked list

Given a singly linked list, determine if it is a palindrome.

This can be solved by reversing the 2nd half and compare the two halves. Let’s start with an example [1, 1, 2, 1].

In the beginning, set two pointers fast and slow starting at the head.
1 -> 1 -> 2 -> 1 -> null
sf
(1) Move: fast pointer goes to the end, and slow goes to the middle.
1 -> 1 -> 2 -> 1 -> null
　　　　s　　　　f
(2) Reverse: the right half is reversed, and slow pointer becomes the 2nd head.
1 -> 1 null <- 2 <- 1
h　　　　　　　s
(3) Compare: run the two pointers head and slow together and compare.
1 -> 1 null <- 2 <- 1
　　h　　　s

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isPalindrome(ListNode head) {
        ListNode fast=head;
        ListNode slow=head;
        while(fast!=null&&fast.next!=null){
            fast=fast.next.next;
            slow=slow.next;
        }
        if(fast!=null){
            slow=slow.next;
        }
        slow=reverse(slow);
        fast=head;
        while(slow!=null){
            if(slow.val!=fast.val){
                return false;
            }
            slow=slow.next;
            fast=fast.next;
        }
        return true;
    }
    public static ListNode reverse(ListNode head){
        ListNode pre=null;
        while(head!=null){
            ListNode temp=head.next;
            head.next=pre;
            pre=head;
            head=temp;
        }
        return pre;
    }
}