Add Two Numbers(LeetCode)

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

这是一个基础的链表问题，需要注意的是初始链表l1,l2可能为NULL的情况，则可直接返回另一链表。

/** * 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 flag = 0;        if( l1 == NULL )            return l2;        else if( l2 == NULL )            return l1;        ListNode *p,*head = new ListNode(0);        p = head;        head -> val = (l1 -> val + l2 -> val) % 10;        flag = (l1 -> val + l2 -> val) / 10;        l1 = l1 -> next;        l2 = l2 -> next;        while( l1 != NULL || l2 != NULL)    {            int num1 = 0,num2 = 0;            int num;            if( l1 != NULL )    {                num1 = l1 -> val;                l1 = l1 -> next;            }            if( l2 != NULL )    {                num2 = l2 -> val;                l2 = l2 -> next;            }            num = ( num1 + num2 + flag ) % 10;            flag = ( num1 + num2 + flag ) / 10;            ListNode *q = new ListNode(num);            p -> next = q;            p = p -> next;        }        if( flag != 0 ) {            ListNode *q = new ListNode(1);            p -> next = q;            p = p -> next;        }        return head;    }};