题目描述：

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

Input: (2 -> 4 -> 3) + (5 -> 6-> 4)

Output: 7 -> 0 -> 8

解法：

本题其实是大数按位相加的变形。其中值得注意的是不要忘了进位位可以当独作为一位数输出。如[5],[5]，相加等于[0,1]。很容易忘了1的存在得出[0]的错误结果。

/**
 * 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) {
        if(l1 ==NULL)
            return l2;
        else if(l2 == NULL)
        return l1;
        ListNode* pHead = new ListNode(0);
        pHead->val = 0;
        pHead->next = NULL;//表头pHead
        ListNode * pNext = pHead;//pNext依序递增链表
       int flag = 0, sum = 0;
        while( l1 != NULL && l2 != NULL)
        {
            sum = l1->val + l2->val + flag;
            flag = sum / 10;
            sum = sum % 10;
            pNext->next = new ListNode(sum);//链接链表
            pNext = pNext ->next;
            l1 = l1->next;
            l2 = l2->next;
        }
        while(l1 != NULL)//我自己很容易写成if(l1 != NULL)，一定要注意l1后面也行还有很多值应该直接循环到末尾
        {
            sum = l1->val + flag;
            flag = sum / 10;
            sum = sum % 10;
            pNext->next = new ListNode(sum);
            pNext = pNext ->next;
            l1 = l1->next;
        }
        while(l2 != NULL)
        {
            sum = l2->val + flag;
            flag = sum / 10;
            sum = sum % 10;
            pNext->next = new ListNode(sum);
            pNext = pNext ->next;
            l2 = l2->next;
        }
        if(flag)//！！！当l1和l2位数相等时，千万不用忘了进位位，进位可以单独作为一个数存在!!!
        pNext->next = new ListNode(1);
      return pHead->next;
    }
};