题解 | 链表相加(二)

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head1 ListNode类 
     * @param head2 ListNode类 
     * @return ListNode类
     */
    ListNode* reverse(ListNode* head){
        ListNode* first=nullptr;
        ListNode* second=head;
        ListNode* third;
        while(second!=nullptr){
            third=second->next;
            second->next=first;
            first=second;
            second=third;
        }
        return first;
    }
    ListNode* addInList(ListNode* head1, ListNode* head2) {
        int tem=0;
        head1=reverse(head1);
        head2=reverse(head2);
        ListNode* nhead=new ListNode(-1);
        ListNode* origin_head=nhead;
        while(head1!=nullptr||head2!=nullptr){
            int val=tem;
            if(head1!=nullptr){
                val+=head1->val;
                head1=head1->next;
            }
            if(head2!=nullptr){
                val+=head2->val;
                head2=head2->next;
            }
            tem=val/10;
            val=val%10;
            nhead->next=new ListNode(val);
            nhead=nhead->next;
            
        }
        if(tem>0){
            nhead->next=new ListNode(tem);
            nhead=nhead->next;
        }
        return reverse(origin_head->next);

    }
};