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
Explanation: 342 + 465 = 807.
代码(C++实现)：
/**
* 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) {
// 结果链表的首节点
ListNode *head = nullptr;
// 尾插法建立单链表指向尾节点
ListNode *tail = nullptr;
// num1为两整数相加后的个位数值
int num1 = ;
// num2为两整数相加后的进位数值
int num2 = ;
// node1Val为l1当前节点的数据域或者0
int node1Val = ;
// node2Val为l2当前节点的数据域或者0
int node2Val = ;
// count变量的设置是为了尾插法建立单链表设置的计数器
int count = ; while(l1 != nullptr || l2 != nullptr)
{
// 如果节点不为空，则取节点的数据域否则让节点的数据域为0
if(l1 == nullptr)
{
node1Val = ;
}else
{
node1Val = l1->val;
}
if(l2 == nullptr)
{
node2Val = ;
}else
{
node2Val = l2->val;
}
// 本次计算结果 = 本次计算的node1Val + 本次计算的node1Va2 + 进位值
num1 = node1Val + node2Val + num2;
if(num1 >= )
{
num1 = num1 - ;
num2 = ;
}else
{
num2 = ;
}
// 为建立结果链表创建节点
ListNode *newNode = new ListNode(num1);
// 尾插法建立结果单链表，如果是首节点，需要进行特殊处理
if(count == )
{
head = tail = newNode;
}else
{
tail->next = newNode;
tail = newNode;
}
// 链表向前移动并释放原始链表所占的内存空间
if(l1 != nullptr)
{
ListNode *tempNode1 = l1;
l1 = l1->next;
delete tempNode1;
}
if(l2 != nullptr)
{
ListNode *tempNode2 = l2;
l2 = l2->next;
delete tempNode2;
} /* 为了解决例如情况：l1 = [5]
l2 = [5]
这种情况
*/
if(l1 == nullptr && l2 == nullptr && num2 != )
{
ListNode *newNode = new ListNode(num2);
tail->next = newNode;
tail = newNode;
return head;
}
count++; } return head;
}
};