[leetcode]Add Two Numbers

纯水题，想来想去指针链表类题目怎么都还是C++比较好。环境里默认没有NULL，就先用0代替了。觉得风格比较烂。

后找到一个更精炼的代码，一是使用二级指针，省了一个前向变量，二是二链表同时前进结束后，做个简单判断，就不用两个while了。三是进位不知道，随便起了个up，其实overflow或是carry都要好多了。

本人的代码：

class Solution {
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
    	int up = 0;
		ListNode * head = 0;
		ListNode * last = 0;
		ListNode * tmp = 0;
		while (l1 != 0 && l2 != 0)
		{
			int val = 0;
			calculateValueAndUp(&val, &up, l1->val, l2->val);
			tmp = new ListNode(val);
			if (head == 0)
			{
				head = tmp;
			}
			else
			{
				last->next = tmp;
			}
			last = tmp;
			l1 = l1->next;
			l2 = l2->next;
		}
        while (l1 != 0)
		{
			int val = 0;
			calculateValueAndUp(&val, &up, l1->val, 0);
			tmp = new ListNode(val);
			if (head == 0)
			{
				head = tmp;
			}
			else
			{
				last->next = tmp;
			}
			last = tmp;
			l1 = l1->next;
		}
		while (l2 != 0)
		{
			int val = 0;
			calculateValueAndUp(&val, &up, 0, l2->val);
			tmp = new ListNode(val);
			if (head == 0)
			{
				head = tmp;
			}
			else
			{
				last->next = tmp;
			}
			last = tmp;
			l2 = l2->next;
		}
		if (up == 1)
		{
			tmp = new ListNode(1);
			last->next = tmp;
		}
        return head;
    }

	void calculateValueAndUp(int* val, int* up, int a, int b)
	{
		*val = a + b + *up;
		if (*val >= 10)
		{
			*val -= 10;
			*up = 1;
		}
        else
        {
            *up = 0;
        }
	}
};

更好的代码

class Solution {
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        int overflow = 0;
        ListNode *ret = NULL;
        ListNode **pnode = &ret;
        while(l1 && l2) {
            int val = l1->val + l2->val + overflow;
            overflow = val / 10;
            *pnode = new ListNode(val % 10);
            pnode = &((*pnode)->next);
            l1 = l1->next;
            l2 = l2->next;
        }
        ListNode *lremain = l1 ? l1 : l2;
        while(lremain) {
            int val = lremain->val + overflow;
            overflow = val / 10;
            *pnode = new ListNode(val % 10);
            pnode = &((*pnode)->next);
            lremain = lremain->next;
        }
        if(overflow > 0) {
            *pnode = new ListNode(overflow);
        }
        return ret;
    }
};