234. Palindrome Linked List

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

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

快慢指针找中点，reverse后半段（slow.next），比较两段。注意找到中点后要断开，否则会无限循环。

/**
 * 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 slow = head;
        ListNode fast = head.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode head2 = slow.next;
        slow.next = null;
        head2 = reverse(head2);
        while (head != null && head2 != null) {
            if (head.val != head2.val) {
                return false;
            }
            head = head.next;
            head2 = head2.next;
        }
        return true;
    }

    private ListNode reverse(ListNode head) {
        ListNode prev = null;
        while (head != null) {
            ListNode temp = head.next;
            head.next = prev;
            prev = head;
            head = temp;
        }
        return prev;
    }
}