LeetCode 2 Add Two Numbers

题目

You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

解法

链表顺序与数字顺序相反，表头存放的是数字的最低位，(2 -> 4 -> 3) + (5 -> 6 -> 4)即342+465=807，可以直接在链表上进行对应位的加法，向后进位。
因为题目中说了是非空链表，所以初始不再考虑l1和l2为空的情况；
l1和l2长度可能不同，对于较长链表剩余的高位，与0相加得到结果；

/** * 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) {        int sum = l1->val + l2->val;        ListNode* l3 = new ListNode(sum%10);        ListNode* node = l3;        l1 = l1->next;        l2 = l2->next;        sum = sum/10;        while( l1!=NULL || l2!=NULL || sum != 0) {            int n1 = 0, n2 = 0;            if (l1!=NULL) {                n1 = l1->val;                l1 = l1->next;            }                if (l2!=NULL) {                n2 = l2->val;                l2 = l2->next;            }            sum+= n1 + n2;            node->next = new ListNode(sum%10);            node = node->next;            sum = sum/10;        }        return l3;    }};