1,先找到中间节点

2.逆置中间节点以后的链表

3.让第一个节点与此时中间节点依次比较

关于单链表问题,欢迎进入我的博客,我会分享学习过程中的感悟 https://blog.csdn.net/m0_63111921/article/details/122463378?spm=1001.2014.3001.5501 alt /* struct ListNode { int val; struct ListNode next; ListNode(int x) : val(x), next(NULL) {} };/ class PalindromeList { public: bool chkPalindrome(ListNode* A) { struct ListNodesnow=A; struct ListNodefast=A; while(fast&&fast->next) { snow=snow->next; fast=fast->next->next; }

struct ListNode*newhead=NULL;
 struct ListNode*cur=snow;
while(cur)
{
    struct ListNode*next=cur->next;
    cur->next=newhead;
    newhead=cur;
    cur=next;
}
    while(A&&newhead)
    {
        if(A->val!=newhead->val)
        {
             return false;
        }
        else
        {
            A=A->next;
            newhead=newhead->next;
        }
    }
    return true;
}

};