[LeetCode] Add Two Numbers - c++编程基础

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

问题描述：给定两个单链表，它们分别代表一个非负的整数，每个节点存储一个数字，将两个单链表代表的整数相加，并返回结果单链表。

这道题没说是在原单链表的基础上进行相加，还是建立一个新的单链表，简单起见，这里采用新建一个单链表来对两个链表进行相加。

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;  
          
        int ain = 0;  
        int tmp = 0;  
          
        ListNode *p = l1, *q = l2;  
        ListNode *r = NULL, *m = NULL, *n = NULL;  
        while(p && q) {  
            tmp = p->




val + q->val + ain;  
            ain = tmp / 10;  
            tmp = tmp % 10;  
            if(r == NULL) {  
                r = new ListNode(tmp);  
                m = r;  
            }  
            else {  
                m->next = new ListNode(tmp);  
                m = m->next;  
            }  
            p = p->next;  
            q = q->next;  
        }  
          
        while(p) {  
            tmp = p->val + ain;  
            ain = tmp / 10;  
            tmp = tmp % 10;  
            m->next = new ListNode(tmp);  
            m = m->next;  
            p = p->next;  
        }  
          
        while(q) {  
            tmp = q->val + ain;  
            ain = tmp / 10;  
            tmp = tmp % 10;  
            m->next = new ListNode(tmp);  
            m = m->next;  
            q = q->next;  
        }  
          
        if(ain) {  
            m->next = new ListNode(ain);  
        }  
          
        return r;  
    }  
};