该题目的要求是判断一个单链表是否是回文链表，题目的难度在于O(n)时间和O(1)空间的限制。

由于单链表不能反向访问，所以不能直接通过原来的链表来判断，解题的思路是首先对原来的链表的前半部分进行判断，然后进行判断（如链表为“12344321” 反转之后为“43214321”）。想到这一点之后的实现就非常简单了，完整的代码如下所示：

?

class Solution {
public:
    ListNode* reverseHalf(ListNode* root){
        ListNode *fast = root, *slow = root;
//slow 指向要反转的部分的后一个节点        
        while(fast != nullptr && fast -> next != nullptr){
            fast = fast -> next -> next;
            slow = slow -> next;
        }
        
        ListNode *h = new ListNode(0);
        h -> next = root;
//执行反转
        ListNode *sec = root, *fir = root -> next;
        while(fir != slow){
            sec -> next = fir -> next;
            fir -> next = h -> next;
            h -> next = fir;
            fir = sec -> next;
        }
        return h -> next;
    }

    bool isPalindrome(ListNode* head) {
        int len = 0;
        ListNode* p = head;
        ListNode *fast, *slow;
//计算单链表的长度
        while(p != NULL){
            len++;
            p = p -> next;
        }
        if(len < 2)
            return true;
//反转单链表的前半部分            
        ListNode* reversedList = reverseHalf(head);
//slow 指向单链表的中间位置
        fast = slow = reversedList;
        while(fast != nullptr && fast -> next != nullptr){
            fast = fast -> next -> next;
            slow = slow -> next;
        }
//fast 指向单链表的头部
        fast = reversedList;
        if(len % 2 != 0)
            slow = slow -> next;
//判断是否为回文子串
        while(slow  != nullptr){
            if(fast -> val != slow -> val)
                return false;
            fast = fast -> next;
            slow = slow -> next;
        }
        return true;
    }
};

?

?