链表的回文结构

1->2->2->1
返回：true

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};*/
class PalindromeList {
public:
    bool chkPalindrome(ListNode* A) {
        ListNode* fast = A;
        ListNode* slow = A;
        //先找到中间节点
        while(fast != nullptr && fast->next != nullptr)
        {
            fast = fast->next->next;
            slow = slow->next;
        }
        
        //然后将中间结点后面的结点逆置
        ListNode* prev = nullptr;
        while(slow != nullptr)
        {
            ListNode* next = slow->next;
            slow->next = prev;
            prev = slow;
            slow = next;
        }
        
        //比较,若相同返回true，不同返回false
        //注意：这里是用pre来和A比，详细的可以自己画图体会
        while(prev != nullptr && A!= nullptr)
        {
            if(prev->val != A->val)
            {
                return false;
            }
            else
            {
                prev = prev->next;
                A = A->next;
            }
        }
        return true;
    }