[LeetCode]Palindrome Linked List

反转后一半，然后判断，不过这样子会改变输入的数据，感觉不太好。也可以用一个栈，但是那样的话空间复杂度就不符合标准了。

/**
 * 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) {
        if (head == null || head.next == null) {
            return true;
        }
        int length = 0;
        ListNode p = head;
        while (p != null) {
            length ++;
            p = p.next;
        }
        p = head;
        ListNode pre = p;
        for (int i = 0; i < length / 2; i++) {
            pre = p;
            p = p.next;
        }
        if (length % 2 == 0) {
            pre.next = null;
            p = reverse(p);
        } else {
            pre.next = null;
            p = p.next;
            p = reverse(p);
        }
        while (head != null) {
            if (head.val != p.val) {
                return false;
            }
            head = head.next;
            p = p.next;
        }
        return true;
    }
    public ListNode reverse(ListNode head) {
        ListNode p2 = head;
        ListNode p1 = null;
        while (p2 != null) {
            ListNode tmp = p2;
            p2 = p2.next;
            tmp.next = p1;
            p1 = tmp;
        }
        return p1;
    }
}