Add Two Numbers.txt

problem：
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

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solution：
struct ListNode
{
  int val;
	ListNode* next;
	ListNode(int v):val(v), next(NULL){}
};

//一、注意为NULL时候的处理，这里使用了lRemain省去了很多重复过程
//二、注意到最后还有进位的时候，需要再新建一个节点存进位值
//三、这里用到指针的指针。往空链表中插入结点时，新插入的结点就是链表的头指针，由于此时会改动头指针，因此必须把头结点设为指向指针的指针
//关于指针的指针：作为指针，把其指向的结点加入到相应的子链表中。例如：
//ListNode **head = NULL;  定义一个指针的指针为头结点
//*head = new ListNode(0); 将新结点new ListNode(0)加入到以head为头结点的子链表中
//head = &((*head)->next); 更新head结点为新节点的next位置

ListNode *addTwoNumbers(ListNode *l1, ListNode *l2)
{
	int carry = 0;
	ListNode* ret = NULL; //记录返回新链表的头节点
	ListNode** pNode = &ret; //指针的指针，每次更新向链表中插入新结点
	while(l1&&l2)
	{
		int s = carry + l1->val + l2->val;
		carry = s / 10;
		*pNode = new ListNode(s%10);
		pNode = &((*pNode)->next);
		l1 = l1->next;
		l2 = l2->next;
	}
	ListNode* lRemain = l1 ? l1 : l2;
	while(lRemain)
	{
		int s = carry + lRemain->val;
		carry = s / 10;
		*pNode = new ListNode(s%10);
		pNode = &((*pNode)->next);
		lRemain = lRemain->next;
	}
	if(carry > 0)
	{
		*pNode = new ListNode(carry);
	}
	return ret;
}