题解 | 链表的回文结构

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};*/
//快慢指针找中间节点
ListNode* FindMid(ListNode* head)
{
ListNode* fast=head;
ListNode* slow=head;
while(fast->next!=NULL&&fast!=NULL)
{
    slow=slow->next;
    fast=fast->next->next;
}
return slow;
}

ListNode* reList(ListNode* head)//三指针反转链表
{
   if(head==NULL)
   return head;
   ListNode* n1=NULL,*n2=head,*n3=head->next;
   while(n2!=NULL)
   {
    n2->next=n1;
    n1=n2;
    n2=n3;
    if(n3!=NULL)
    n3=n3->next;
   }
   return n1;
}

class PalindromeList {
public:
    bool chkPalindrome(ListNode* A) {
       ListNode* mid=FindMid(A);//找中间节点
       ListNode* newhead=reList(mid);//反转后新节点
       while(newhead!=NULL)
       {
        if(newhead->val!=A->val)
        return false;
        else
        {
            newhead=newhead->next;
            A=A->next;
        }
       }
       return true;
    }
};