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?

判断一个单向链表是否为回文链表，要求时间复杂度为O(n)，空间复杂度为O(1)。我们可以先用快慢指针找到中间的节点，然后将后半部分链表反转，然后以前半部分的链表进行比较，如果相同就返回true，如果不同就返回false。代码如下：

/**
 * 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;
        ListNode fast = head;
        ListNode slow = head;
        while(fast.next != null && fast.next.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        ListNode newHead = slow.next;
        slow.next = null;
        newHead = invert(newHead);
        while(newHead != null) {
            if(head.val != newHead.val)
                return false;
            head = head.next;
            newHead = newHead.next;
        }
        return true;
    }
    public ListNode invert(ListNode head) {
        ListNode pre = null;
        ListNode tem = null;
        while(head != null) {
            tem = head.next;
            head.next = pre;
            pre = head;
            head = tem;
        }
        return pre;
    }
}