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){
        if(!head) return head;
        ListNode *pre=NULL;
        ListNode *cur=head;
        while(cur){
            ListNode *tmp=cur->next;
            cur->next=pre;
            pre=cur;
            cur=tmp;
        }
        return pre;
    }
    ListNode* addInList(ListNode* head1, ListNode* head2) {
        // write code here
        if(!head1) return head2;
        if(!head2) return head1;
        head1=reverse(head1);
        head2=reverse(head2);
        ListNode* head = new ListNode(0);
        ListNode* cur = head;
        int uper = 0;
        while(head1 || head2){
            int val=uper;
            if(head1) {
                val+=head1->val;
                head1=head1->next;
            }
            if(head2) {
                val+=head2->val;
                head2=head2->next;
            }
            uper=val/10;
            cur->next=new ListNode(val%10);
            cur=cur->next;
        }
        if(uper!=0){
            cur->next=new ListNode(uper);
        }
        return reverse(head->next);


    }
};