[leetcode] Add Two Numbers

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

把两个链表的非负数字相加，如果超过10则需要进位。

分成三步处理：

• 处理两个链表相同长度的部分。
• 处理长链表的部分。
• 对长链表最高位进位的处理。

代码如下：

/**
 * 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) {
            // Start typing your C/C++ solution below
            // DO NOT write int main() function
            ListNode * p= l1; ListNode *q=l2;
            ListNode * k =new ListNode(0);
            ListNode * r=k;
            int remain=0;
            for(;p!=NULL&&q!=NULL; p=p->next,q=q->next)
            {    
                int sss;
                sss=(p->val+q->val+remain);
                r->next =new ListNode (sss%10);
                r=r->next;
                remain=sss/10;
            }
            if(p!=NULL)
            {
                for(;p!=NULL; p=p->next)
                {    
                    int sss;
                    sss=(p->val + remain);
                    r->next =new ListNode (sss%10);
                    r=r->next;
                    remain=sss/10;
                }
            }
            if(q!=NULL)
            {
                for(;q!=NULL; q=q->next)
                {    
                    int sss;
                    sss=(q->val + remain);
                    r->next =new ListNode (sss%10);
                    r=r->next;
                    remain=sss/10;
                }
            }
            if(remain!=0) 
            {
                r->next =new ListNode (1);
                //  k->next=NULL;
            }
            return k->next;
        }
};