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
分析：
这个题目涉及到链表知识，关于指针、链表自从接触就是我的弱势，到现在那么多年过去了，依然没有改变。但是，既然撞上了，还是必须义无反顾的应战的。分析一下题目，它是说有两个链表，每个链表存储非负数值，链表的每个节点都是一位个位数字，数值倒序存储在链表中，两者求和并以链表的形式返回。
题目信息分析完毕后，开始动手吧，哎，依然是头皮发麻呀~
AC代码：
这个题目的代码编写到调试，再到最后的AC，可真是花费了不少功夫呀，好在功夫不负有心人，嘿嘿，问题代码就不上传了，下面贴出AC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; vector<int> v1;
vector<int> v2;
ListNode *head=NULL , *rear=NULL;
while(l1 != NULL)
{
v1.push_back(l1->val);
l1 = l1->next;
}
while(l2 != NULL)
{
v2.push_back(l2->val);
l2 = l2->next;
} if(v1.size() < v2.size())
{
for(int k=v1.size() ; k<v2.size() ; k++)
v1.push_back(0);
}else
{
for(int k=v2.size() ; k<v1.size() ; k++)
v2.push_back(0);
}
int temp = 0;
int value = 0;
for(int j=0 ; j<v1.size() ; j++)
{
int sum = v1[j] + v2[j] + temp;
temp = sum / 10;
value = sum % 10;
ListNode *node = new ListNode(value);
if(head == NULL)
head = node;
if(rear == NULL)
rear = node;
else
{
rear->next = node;
rear = rear->next;
}
}
if(temp != 0 && rear!=NULL)
{
ListNode *node = new ListNode(temp);
rear->next = node;
}
return head;
}
};
测试Main函数：
为了方便测试，下面提供Main测试代码，说明，在LeetCode页面提交代码，只需要上传Solution类即可：
int main()
{
ListNode *l1=NULL , *r1=NULL, *l2 = NULL , *r2=NULL , *result=NULL;
int arr1[3] = {2,4,3};
int arr2[3] = {5,6,4};
for(int i=0 ; i<3 ; i++)
{
ListNode *node1 = new ListNode(arr1[i]);
ListNode *node2 = new ListNode(arr2[i]);
if(l1 == NULL)
l1 = node1;
if(r1 == NULL)
r1 = node1;
else{
r1->next = node1;
r1 = r1->next;
}
if(l2 == NULL)
l2 = node2;
if(r2 == NULL)
r2 = node2;
else{
r2->next = node2;
r2 = r2->next;
}
}
Solution s;
result = s.addTwoNumbers(l1,l2);
for( ; result!=NULL ; result=result->next)
cout<<result->val<<"->";
cout<<endl;
system("pause");
return 0;
}