两个链表相加求和

你有两个用链表代表的整数，其中每个节点包含一个数字。数字存储按照在原来整数中相反的顺序，使得第一个数字位于链表的开头。写出一个函数将两个整数相加，返回和。 

若链表存储是从高位开始存储，则将链表先反转再相加。反转链表实现

#include
#include
#include

struct Node
{
    int data;
    Node* next;
    
    Node() : data(0), next(NULL) {}
};

Node* createList(int* list, int size)
{
    if (0 == size)
    {
        return NULL;
    }
    Node* head = NULL, *p = NULL;
    head = p = new Node();
    p->data = list[0];
    p->next = NULL;
    for (int i = 1; i < size; i++)
    {
        Node* cur = new Node();
        cur->data = list[i];
        cur->next = NULL;
        p->next = cur;
        p = cur;
    }

    return head;
}

void toString(Node* pList)
{
    Node* p = pList;
    while (p)
    {
        std::cout << p->data << ",";
        p = p->next;
    }
    std::cout << std::endl;
}

std::string sumList(Node* pList1, Node* pList2)
{
    int flag = 0;
    Node* p1 = pList1;
    Node* p2 = pList2;
    std::vector result;
    while (true)
    {
        int sum = 0;
        if (p1 != NULL)
        {
            sum += p1->data;
            p1 = p1->next;
        }
        if (p2 != NULL)
        {
            sum += p2->data;
            p2 = p2->next;
        }
        sum += flag;
        flag = sum / 10;
        result.push_back(sum % 10);
        if (p1 == NULL && p2 == NULL)
        {
            break;
        }
    }
    if (flag > 0)
    {
        result.push_back(flag);
    }
    std::ostringstream os;

    for (std::vector::reverse_iterator numIter = result.rbegin(); numIter != result.rend(); ++numIter)
    {
        os << *numIter;
    }
    return os.str();

}

int main()
{
    int list1[] = {1,3,9,8};
    int list2[] = {2,5,7,4,2};

    Node* pList1 = createList(list1, sizeof(list1)/sizeof(int));
    Node* pList2 = createList(list2, sizeof(list2)/sizeof(int));
    toString(pList1);
    toString(pList2);
    std::cout << sumList(pList1, pList2) << std::endl;
    return 0;

}