LeetCode OJ 02 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

难度： medium

解题思路：

注意链表的数字顺序是与实际数字相反，则假如l1和l2的长度不一致，短的那个链表的尾部在做加法时相当于0。

代码

class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        ListNode result(-1);//头节点        int carry = 0;        ListNode *prev = &result;        for(ListNode *pa=l1,*pb=l2;            pa!=nullptr||pb!=nullptr;            pa=pa==nullptr?nullptr:pa->next,            pb=pb==nullptr?nullptr:pb->next,            prev=prev->next){                const int ai=pa==nullptr?0:pa->val;                const int bi=pb==nullptr?0:pb->val;                const int value = (ai+bi+carry)%10;                carry = (ai+bi+carry)/10;                prev->next = new ListNode(value);//尾插法            }            if(carry>0)                prev->next = new ListNode(carry);            return result.next;        }};