输入一个单向链表，输出该链表中倒数第 k 个结点

class ListNode
{
public:
	ListNode()
	{
		pNext	= NULL;
		nValue  = 0;
	}
	ListNode* pNext;
	int nValue;
};
ListNode* CreaList()
{
	int nValue;
	ListNode* Head = NULL;
	ListNode* ListIndex = NULL;
	while(cin >> nValue)
	{
		if (Head == NULL)
		{
			Head = new ListNode();
			Head->nValue = nValue;
			ListIndex = Head;
		}
		else
		{
			ListNode* newNode = new ListNode();
			newNode->nValue = nValue;
			ListIndex->pNext = newNode;
			ListIndex = newNode;
		}
	}

	//
	cout << "Print List;" << endl;
	PrintList(Head);

	return Head;
}

void PrintList(ListNode* Head)
{
	for (ListNode* pNode = Head; pNode != NULL; pNode = pNode->pNext)
	{
		cout << pNode->nValue << " ";
	}
	cout << endl;
}

//运用两个节点 可以达到通过一次遍历完成输出倒数第K个节点
void PrintTheLastK(ListNode* Head, int K)
{
	if (Head == NULL)
	{
		cout << "number of list less then " << K << endl;
		return;
	}

	ListNode*  firstPoint = Head;
	ListNode*  secondPoint = Head;

	for (int i = 1; i < K ; i++)
	{
		secondPoint = secondPoint->pNext;
		if (secondPoint == NULL)
		{
			cout << "number of list less then " << K << endl;
			return;
		}
	}

	while(1)
	{
		if (secondPoint->pNext == NULL)
		{
			cout << "result:" << firstPoint->nValue << endl;
			return;
		}
		firstPoint = firstPoint->pNext;
		secondPoint = secondPoint->pNext;
	}
}

//拓展
//输入一个单向链表。如果该链表的结点数为奇数，输出中间的结点；如果链表结点数为偶数，输出中间两个结点前面的一个
void PrintMidNode(ListNode* Head)
{
	if (Head == NULL)
	{
		return;
	}

	ListNode* firstNode = Head;
	ListNode* secondNode = Head;

	while (1)
	{
		if (secondNode->pNext == NULL || secondNode->pNext->pNext == NULL)
		{
			cout << "result:" << firstNode->nValue << endl;
			return;
		}

		secondNode = secondNode->pNext->pNext;
		firstNode  = firstNode->pNext;
	}
}

输入一个单向链表，输出该链表中倒数第 k 个结点

原文：http://blog.csdn.net/king__moving/article/details/48028513