[LeetCode]Add Two Numbers

题目要求：

You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

这道题其实是大数相加的处理，没什么难度，但需要注意以下几点：

1.因为存储是反过来的，即数字342存成2->4->3，所以要注意进位是向后的；

2.链表l1或l2为空时，直接返回，这是边界条件，省掉多余的操作；

3.链表l1和l2长度可能不同，因此要注意处理某个链表剩余的高位；

4.2个数相加，可能会产生最高位的进位，因此要注意在完成以上1－3的操作后，判断进位是否为0，不为0则需要增加结点存储最高位的进位。

我的代码如下，欢迎大牛指导交流~

AC，Runtime: 216 ms

//LeetCode_Add Two Numbers//Written by zhou//2013.11.1/** * 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) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.                if (l1 == NULL) return l2;if (l2 == NULL) return l1;        ListNode *resList = NULL, *pNode = NULL, *pNext = NULL;        ListNode *p = l1, *q = l2;        int up = 0;        while(p != NULL && q != NULL)        {            pNext = new ListNode(p->val + q->val + up);            up = pNext->val / 10;    //计算进位            pNext->val = pNext->val % 10;   //计算该位的数字                        if (resList == NULL)  //头结点为空            {                resList = pNode = pNext;            }            else //头结点不为空            {                pNode->next = pNext;                pNode = pNext;            }            p = p->next;            q = q->next;        }//处理链表l1剩余的高位while (p != NULL){pNext = new ListNode(p->val + up);up = pNext->val / 10;                pNext->val = pNext->val % 10;pNode->next = pNext;            pNode = pNext;p = p->next;}//处理链表l2剩余的高位while (q != NULL){pNext = new ListNode(q->val + up);up = pNext->val / 10;                pNext->val = pNext->val % 10;pNode->next = pNext;            pNode = pNext;q = q->next;}//如果有最高处的进位，需要增加结点存储if (up > 0){pNext = new ListNode(up);pNode->next = pNext;}        return resList;    }};