判断链表是否为回文结构

对于一个链表，请设计一个时间复杂度为O(n),额外空间复杂度为O(1)的算法，判断其是否为回文结构。

回文结构示例：

1->2->2->1
返回：true

方法实现：

public class PalindromeList {
    public boolean chkPalindrome(ListNode A) {
        // write code here
        if(A == null || A.next == null) {
            return true;
        }
        // 当前链表至少有两个节点
        ListNode fast = A;
        ListNode low = A;
        while(fast!=null&&low!=null&&fast.next!=null) {
            fast = fast.next.next;
            low = low.next;
        }
        // 找到链表的中间位置
        ListNode mid = low;
        // 反转后半段链表
        ListNode dummyHead = new ListNode(-1);
        dummyHead.next = mid;
        if(mid.val == A.val) {
            return true;
        }else {
            if(mid.next == null) {
                return false;
            }else {
                ListNode f = dummyHead.next;
                ListNode s = f.next;
                while(s!=null) {
                    f.next = s.next;
                    s.next = dummyHead.next;
                    dummyHead.next = s;
                    s = f.next;
                }
                // 两个链表开始从前向后遍历
                fast = A;
                while(A!=null&&dummyHead.next!=null) {
                    if(A.val != dummyHead.next.val) {
                        return false;
                    }
                    A = A.next;
                    dummyHead.next = dummyHead.next.next;
                }
                return true;
            }
        }
    }
}