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

Example 1:

Input: 1->2
Output: false

Example 2:

Input: 1->2->2->1
Output: true

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

找中點，反轉後半部分，再一一比較

time: O(n), space: O(1)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 
 
*/
class Solution {
    public boolean isPalindrome(ListNode head) {
        if(head == null || head.next == null) {
            return true;
        }
        ListNode fast = head, slow = head;
        while(fast.next != null && fast.next.next != null) {
            fast = fast.next.next;
            slow 
 
= slow.next;
        }
        ListNode p1 = head, p2 = reverse(slow.next);
        slow.next = null;
        while(p1 != null && p2 != null) {
            if(p1.val != p2.val) {
                return false;
            }
            p1 = p1.next;
            p2 = p2.next;
        }
        
 
return true;
    }
    
    private ListNode reverse(ListNode head) {
        if(head == null || head.next == null) {
            return head;
        }
        ListNode prev = null, cur = head;
        while(cur != null) {
            ListNode next = cur.next;
            cur.next = prev;
            prev = cur;
            cur = next;
        }
        return prev;
    }
}