Sum the linked lists

Sum the linked lists
1->2->3
4->5

result is 1->6->8

 

1->5->9

5->3

result  2->1->2

A: 把链表的值加起来，注意，根据sample，是从链表的最后一位开始相加，而且，要考虑进位的问题

 

pseudo code如下 （通过编译，但没有测试用例）

LinkNode* SumLinkLists(LinkNode *p1, LinkNode *p2)
{
    if (NULL == p1 && NULL == p2)
        return NULL;

    if (NULL == p1 && NULL != p2)
        return p2;

    if (NULL != p1 && NULL == p2)
        return p1;

    std::stack<LinkNode*> s1;
    std::stack<LinkNode*> s2;

    LinkNode* t1 = p1;
    LinkNode* t2 = p2;

    while (t1)
    {
        s1.push(t1);    
        t1 = t1->next;
    }

    while (t2)
    {
        s2.push(t2);
        t2 = t2->next;
    }

    int sum = 0, carry = 0;
    LinkNode* p = new LinkNode();
    LinkNode* h = p;

    while (!s1.empty() && !s2.empty())
    {
        t1 = s1.top();
        t2 = s2.top();

        sum = t1->data + t2->data + carry;    // 得到当前结点的和，同时加上进位

        if (sum > 10)    // 计算需要存储到结点上的'和'及'进位'
        {
            sum = sum%10;
            carry = sum/10;
        }

        LinkNode* n = new LinkNode();
        n->data = sum;

        p->next = n;
        p = p->next;        

        s1.pop();
        s2.pop();
    }

    // 将剩余的数据也添加进去
    std::stack<LinkNode*> &refL = (!s1.empty())? s1:s2;
    while (!refL.empty())
    {
        sum = refL.top()->data + carry;
        if (sum > 10)
        {
            sum = sum%10;
            carry = sum/10;
        }

        LinkNode* n = new LinkNode();
        n->data = sum;
        p->next = n;
        p = p->next;
        refL.pop();
    }

    p = h->next; // 得到真正的head
    delete h;
    h = NULL;

    // reverse the link list p
    // 反转链表
    LinkNode* pre = p;
    p = p->next;
    pre->next = NULL;

    while (p)
    {
        // pre -> p -> nex
        LinkNode* n = p->next;

        p->next = pre;
        pre = p;
        p = n;
    }

    return pre;
}