题解 | #链表的回文结构#

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};*/
class PalindromeList {
public:
    bool chkPalindrome(ListNode* A) 
    {
        //思路2:利用快慢指针找到中间节点，然后逆置中间节点以及往后的节点，最后判断两个链表每个节点是否相等
        struct ListNode*slow,*fast;
        slow = fast = A;
        while(fast && fast->next)
        {
            slow = slow->next;
            fast = fast->next->next;
        } 
        struct ListNode*tail = NULL;
        struct ListNode*cur = slow;
        while(cur)
        {
            cur = cur->next;
            slow->next = tail;
            tail = slow;
            slow = cur;
        }
        while(tail && A)
        {
            if(tail->val == A->val)
            {
                tail = tail->next;
                A = A->next;
            }
            else 
            {
                return false;;
            }
        }
        return true;
    }
};