链表相加(二) 2022.5.13

class Solution {
public:
  
    ListNode* ReverseList(ListNode* head)
    {
        ListNode* cur=head;
        ListNode* pre=NULL;
        if(head==NULL)
        {return head;}  
        while(cur!=NULL)
        {
            ListNode* tmp=cur->next;
            cur->next=pre;
            pre=cur;
            cur=tmp;
        }
        return pre;
    }
    ListNode* addInList(ListNode* head1, ListNode* head2) {
        if(head1==nullptr)
            return head2;
        if(head2==nullptr)
            return head1;
        
        
        //反转两个链表
        head1=ReverseList(head1);
        head2=ReverseList(head2);
        
        ListNode* head=new ListNode(-1);
        ListNode* ptr=head;
        
        int tmp=0;
        while(head1!=nullptr||head2!=nullptr)
        {
            int val=tmp;
            if(head1!=nullptr)
            {
                val+=head1->val;
                head1=head1->next;
            }
              if(head2!=nullptr)
            {
                val+=head2->val;
                head2=head2->next;
              }
            
            //算出进位
            tmp=val/10;
            //算出余位
            ptr->next=new ListNode(val%10);
            ptr=ptr->next;
           
            
        }
         //链表加完后，注意最后一次的进位
        if(tmp>0)
            {
                ptr->next=new ListNode(tmp);
            }   
        
          return ReverseList(head->next);
    }
};