題目要求：

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

這道題其實是大數相加的處理，沒什麼難度，但需要注意以下幾點：

1.因為儲存是反過來的，即數字342存成2->4->3，所以要注意進位是向後的；

2.連結串列l1或l2為空時，直接返回，這是邊界條件，省掉多餘的操作；

3.連結串列l1和l2長度可能不同，因此要注意處理某個連結串列剩餘的高位；

4.2個數相加，可能會產生最高位的進位，因此要注意在完成以上1－3的操作後，判斷進位是否為0，不為0則需要增加結點儲存最高位的進位。

我的程式碼如下，歡迎大牛指導交流~

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;    //計算進位
            pNext->val = pNext->val % 10;   //計算該位的數字
            
            if (resList == NULL)  //頭結點為空
            {
                resList = pNode = pNext;
            }
            else //頭結點不為空
            {
                pNode->next = pNext;
                pNode = pNext;
            }
            p = p->next;
            q = q->next;
        }

		//處理連結串列l1剩餘的高位
		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;
		}

		//處理連結串列l2剩餘的高位
		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;
		}

		//如果有最高處的進位，需要增加結點儲存
		if (up > 0)
		{
			pNext = new ListNode(up);
			pNode->next = pNext;
		}

        return resList;
    }

};