代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
    ListNode dummy (0);
    ListNode* p = &dummy;
    int sum = 0;
    while(l1 || l2 || sum)
    {
        if(l1) 
        { 
            sum += l1->val;
            l1 = l1->next;
        }
        if(l2)  
        {
            sum += l2->val;
            l2 = l2->next;
        }
        p->next = new ListNode(sum % 10);           //sum % 10为当前位的值
        p = p->next;
        sum /= 10;                              //sum/10为进位的值，并用做下一次求和的初值
    }
    return dummy.next;
    }
};