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 /* 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;
}
};