题解 | #链表的回文结构#

class PalindromeList {
public:
    struct ListNode* prevnode(struct ListNode*head)
    {
        struct ListNode*n1,*n2,*n3;
        n1 = nullptr;
        n2 = head;
        n3 = head->next;
        while(n2)
        {
            n2->next = n1;
            n1 = n2;
            n2 = n3;
            if(n3)
            n3 = n3->next;
        }
        return n1;//返回逆置后的链表头节点
    }

    struct ListNode* Midnode(struct ListNode *phead)
    {
         struct ListNode*quick,*slow;
         quick = slow = phead;
         while(quick && quick->next)
         {
            quick = quick->next->next;
            slow  =slow->next;
         }
         return slow;//返回中间节点
    }

    bool chkPalindrome(ListNode* A) {
       struct ListNode*Mid = Midnode(A);
       struct ListNode*pMid  = prevnode(Mid);

       while(A && pMid)
       {
          if(A->val != pMid->val)
          {
              return false;
          }
          A = A->next;
          pMid = pMid->next;
       }
       return true;
    }
};