[LeetCode OJ][002]Add Two Numbers

给定两个表示非负数的链表。数字倒序储存在节点中且每个节点只含一个数字。将两个数相加并返回和的链表。
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

我的想法：
* 两个链表都是非空的，即至少含有一个数字
* 依次相加每个数字，当一个链表便利到尾时直接复制另一个链表。每一次储存结构都要考虑是否进位
* 题目不允许用C，因为提供了构造函数，C++应该是要求用new新建节点

C++03版（55ms）：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution
{
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2)
    {
        ListNode *answer, *current;
        bool carry = false;

        answer = current = new ListNode(l1->val + l2->val);
        l1 = l1->next;
        l2 = l2->next;
        if(current->val >= 10)
        {
            current->val -= 10;
            carry = true;
        }
        while(l1 && l2)
        {
            if(carry)
            {
                current = current->next = new ListNode(l1->val + l2->val + 1);
            }
            else
            {
                current = current->next = new ListNode(l1->val + l2->val);
            }
            l1 = l1->next;
            l2 = l2->next;
            if(current->val >= 10)
            {
                current->val -= 10;
                carry = true;
            }
            else
            {
                carry = false;
            }
        }
        while(l1)
        {
            if(carry)
            {
                current = current->next = new ListNode(l1->val + 1);
            }
            else
            {
                current = current->next = new ListNode(l1->val);
            }
            l1 = l1->next;
            if(current->val >= 10)
            {
                current->val -= 10;
                carry = true;
            }
            else
            {
                carry = false;
            }
        }
        while(l2)
        {
            if(carry)
            {
                current = current->next = new ListNode(l2->val + 1);
            }
            else
            {
                current = current->next = new ListNode(l2->val);
            }
            l2 = l2->next;
            if(current->val >= 10)
            {
                current->val -= 10;
                carry = true;
            }
            else
            {
                carry = false;
            }
        }
        if(carry)
        {
            current->next = new ListNode(1);
        }
        return answer;
    }
};