题解 | #链表相加(二)#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
class Solution {
public:

    ListNode* RverseList(ListNode* pHead) {
        ListNode* next = nullptr;
        ListNode* prev = nullptr;
        ListNode* curr = pHead;
        // ListNode* next = nullptr;

        if(pHead == nullptr) {
            return nullptr;
        }

        while(curr != nullptr) {
            next = curr->next;
            curr->next = prev;
            prev = curr;
            curr = next;  // 收尾呼应
        }
        
        return prev;
    }

    /**
     * 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
        if(head1 == nullptr) {
            return head2;
        }

        if(head2 == nullptr) {
            return head1;
        }

        ListNode* rv1 = RverseList(head1);
        ListNode* rv2 = RverseList(head2);

        int carry = 0;

        ListNode* res = new ListNode(-1);
        ListNode* curr = res;  // 必须用头结点拷贝才能操作next

        while(rv1 != nullptr || rv2 != nullptr || carry != 0){
            
            int val1 = rv1 != nullptr ? rv1->val : 0;
            int val2 = rv2 != nullptr ? rv2->val : 0;

            int temp = val2 + val1 + carry;

            carry = temp / 10;
            temp = temp % 10;

            curr->next = new ListNode(temp);
            curr = curr->next;

            if(rv1 != nullptr) {
                rv1 = rv1->next;
            }

            if(rv2 != nullptr) {
                rv2 = rv2->next;
            }

        }

        return RverseList(res->next);

    }
};