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

这是考察单链表的相关操作，主要就是插入与边界的检测问题。我的想法是：

1.规定用 l1 来返回这个结果；

2.因为要遍历 l1与l2 的所有结点，所以定义两个遍历指针,分别为 p1, p2,同时，由于存在一个链长一个短的情况，为了方便程序的编写，添加一个前置指针，用于添加新的结点或者连接 l1 与 l2

3.p1与 p2 指向的结点有四种情况，分别是：都为null, 只有一个为null, 都非null.于是分四种情况。程序如下：

/** * 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) {        //把两条链合成一条l1，返回        ListNode *p1, *p2, *pre;        int sum = 0;int add = 0;        p1 = l1;        p2 = l2;        pre = p1;if (!l1)return l2;if (!l2)return l1;        while ( p1 && p2 ){sum = p1->val + p2->val + add;p1->val = sum % 10;add = sum / 10;pre = p1;p1 = p1->next;p2 = p2->next;}if ( !p1 && !p2 ){if (add){ListNode* node = new ListNode(1);pre->next = node;}}else if (p1)//p1 不为空{while (p1){sum = p1->val + add;p1->val = sum % 10;add = sum / 10;pre = p1;p1 = p1->next;}if (add){ListNode* node = new ListNode(1);pre->next = node;}}else//p2不为空{pre->next = p2;while (p2){sum = p2->val + add;p2->val = sum % 10;add = sum / 10;pre = p2;p2 = p2->next;}if (add){ListNode* node = new ListNode(1);pre->next = node;}}return l1;    }};