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* addInList(ListNode* head1, ListNode* head2) {
        // write code here
        head1 = ReverseList(head1);
        head2 = ReverseList(head2);
        ListNode LL = ListNode(-1);
        ListNode* RetPhead = ≪
        ListNode* ret  = RetPhead;

        int Carry = 0;
        int h1 = 0;
        int h2 = 0;

        while (head1 || head2) {
            h1 = head1 == nullptr ? 0 : head1->val;
            h2 = head2 == nullptr ? 0 : head2->val;

            ret->next = new ListNode( (h1 + h2 + Carry) % 10 );

            Carry = (h1 + h2 + Carry) / 10;
       
            ret = ret->next;
         
            if(head1) head1 = head1->next;
            if(head2) head2 = head2->next;

        }
        if(Carry){

            ret->next = new ListNode(1);
            ret = ret->next;
        }

        return ReverseList(RetPhead->next);


    }



     ListNode* ReverseList(ListNode* head) {
        if(!head || !head->next) return head;

        ListNode* Pre = nullptr,*Phead = head,*temp = nullptr;
        while (Phead) {
            temp = Phead->next;
            Phead->next = Pre;
            Pre = Phead;
            Phead = temp;

        }
        return Pre;
    
    }
    
};