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

问题描述：给定两个链表，每个链表当中代表两个非负数字，根据输入和输出可以看到

输入：   2   4   3

         5   6   4

输出：   7   0   8

可以看出根据链表依次进行相加小于10，则直接添加在新建列表中，如果大于10，则存储进位值，作为下一节点的值进行添加

class Solution {
private:
	ListNode* ret;
public:
	ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
		ListNode* head_1 = l1,*head_2=l2;
		int index = 0;
		int value = 0;
		int high_value = 0;
		 ret= new ListNode(0);
		ListNode* last_ret = ret;
		while (head_1 != NULL&&head_2 != NULL)
		{
			value = head_1->val + head_2->val+high_value;
			if (value < 10)
			{
				high_value = 0;
				last_ret->val = value;
			}
			else
			{
				high_value = value / 10;
				value = value % 10;
				last_ret->val = value;
			}
			head_1 = head_1->next;
			head_2 = head_2->next;
			if (head_1 != NULL&&head_2 != NULL)
			{
				last_ret->next = new ListNode(0);
				last_ret = last_ret->next;
			}
		}
		ListNode* cur = NULL;
		if (head_1 != NULL)
		{
			cur = head_1;
		}
		else
		{
			cur = head_2;
		}
		if (cur != NULL)
		{

			last_ret->next = new ListNode(0);
			last_ret = last_ret->next;
			while (cur != NULL)
			{
				value = cur->val + high_value;
				if (value < 10)
				{
					high_value = 0;
				}
				else
				{
					high_value = value / 10;
					value = value % 10;
				}
				last_ret->val = value;
				cur = cur->next;
				if (cur != NULL)
				{
					last_ret->next = new ListNode(0);
					last_ret = last_ret->next;
				}

			}			
		}
		if (high_value > 0)
		{
			last_ret->next = new ListNode(0);
			last_ret = last_ret->next;
			last_ret->val = high_value;
		}
		return ret;
	}
	~Solution()
	{
		ListNode* cur = NULL;
		cur = ret;
		while (cur != NULL)
		{
			ListNode* m = cur;
			if (cur != NULL)
			{
				cur = cur->next;
			}
			if(m)
			delete m;
		}
	}
};