老师给出了使用链表和输入数字作为字符串添加 2 个带符号的很长整数的任务我使用的算法，我可以使用其他更好的算法吗？

如何解决老师给出了使用链表和输入数字作为字符串添加 2 个带符号的很长整数的任务我使用的算法，我可以使用其他更好的算法吗？

首先我创建了结构来创建链表。然后创建类和适当的函数来执行某些任务。在时间复杂度的情况下，我可以使用哪些其他方法或算法来使我的解决方案更高效、更快速？

我的解决方案使用了双向链表。我可以使用单个或循环链表来做到这一点吗？

class GiantInt
{
    public:
        node* head;
        node* tail;
        int size; 
        GiantInt()
        {
            head = tail = NULL;
        }
        //~GiantInt();

        void Insert_at_head(int value)
        {
            node* temp = new node(value);

            if(head == NULL)
            {
                head = tail = temp;
            }
            else
            {
                head->prevIoUs = temp;
                temp->next = head;
                head = temp;
            }

            size++;
        }

        void Insert_at_tail(int value)
        {
            node* temp = new node(value);

            if(tail == NULL)
            {
                head = tail = temp;
            }
            else
            {
                tail->next = temp;
                temp->prevIoUs = tail;
                tail = temp;
            }

            size++;
        }

        void display()
        {
            node* temp = head;

            while(temp != NULL)
            {
                cout << temp->data;
                temp = temp->next;
            }
        }

        int length()
        {
            return size;
        }

        void Addition(GiantInt* a,GiantInt* b)
        {
            int c = 0,s;
            GiantInt* a1 = new GiantInt(*a);
            GiantInt* b1 = new GiantInt(*b);

            this->head = NULL;
            this->tail = NULL;
            this->size = 0;

            while (a1->tail != NULL || b1->tail != NULL)
            {
                if(a1->tail != NULL && b1->tail != NULL)
                {
                    s = ((a1->tail->data) + (b1->tail->data) + c) % 10;
                    c = ((a1->tail->data) + (b1->tail->data) + c) / 10;
                    a1->tail = a1->tail->prevIoUs;
                    b1->tail = b1->tail->prevIoUs;    
                }
                else if(a1->tail == NULL && b1->tail != NULL)
                {
                    s = ((b1->tail->data) + c) % 10;
                    c = ((b1->tail->data) + c) / 10;
                    b1->tail = b1->tail->prevIoUs;
                }
                else if(a1-> tail != NULL && b1->tail == NULL)
                {
                    s = ((a1->tail->data) + c) % 10;
                    c = ((a1->tail->data) + c) / 10;
                    a1->tail = a1->tail->prevIoUs;
                }
                Insert_at_head(s);
            }

            if( c != 0)
                Insert_at_head(c);
        }
};

int main()
{
    GiantInt* m = new GiantInt();
    GiantInt* n = new GiantInt();
    GiantInt* sum = new GiantInt();
    
    string number1,number2;
    
    cout<<" Enter numbers to add: ";
    cin >> number1 >> number2;

    for(int i = 0; i < number1.length(); i++)
    {
        m->Insert_at_tail(number1.at(i) - '0');
    }

    for(int i = 0; i < number2.length(); i++)
    {
        n->Insert_at_tail(number2.at(i) - '0');
    }

    cout << "Sum: " << endl;
    sum->Addition(m,n);
    sum->display();
    cout << endl;

    return 0;
}