LeetCode - 2. Add Two Numbers

题目：

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

题目大意：

给你两个链表，链表中存放的是非负数。链表中的数字是逆序存放，并且每个节点都只有一位数字。求讲这两个数字相加得到的链表

考点：

1. 链表操作
2. n位数相加，考虑进位
3. 两个数相加，可能位数不同
4. 如果最后还有一个进位，需要再添加一个节点保存进位数字

代码：

/** * 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) {        if(l1 == NULL) return l2;        if(l2 == NULL) return l1;        ListNode* p1 = l1;        ListNode* p2 = l2;        ListNode* result = NULL;        ListNode* pResult = NULL;        int jinwei = 0;        while(p1 !=NULL || p2 != NULL)        {            int num1 = (p1 == NULL ? 0 : p1->val);            int num2 = (p2 == NULL ? 0 : p2->val);            int val = (num1 + num2 + jinwei) % 10;            jinwei = (num1 + num2 + jinwei) / 10;            if(result == NULL)            {                result = new ListNode(val);                pResult = result;            }            else            {                ListNode* temp = new ListNode(val);                pResult->next = temp;                pResult = pResult->next;            }            p1 = (p1 == NULL? NULL: p1->next);            p2 = (p2 == NULL? NULL: p2->next);        }        if( jinwei > 0){            ListNode* temp = new ListNode(jinwei);            pResult->next = temp;            pResult = temp->next;        }        return result;    }};