解析：设立两个指针，p每次移动两下，q每次只移动一下，那么当p指向最后一个节点的时候，那么q就是中间的节点了

ListNode* FindMidNode(ListNode* pHead){	if (pHead == NULL)	{		return NULL;	}	if (pHead->m_pNext == NULL || pHead->m_pNext->m_pNext == NULL)	{		return pHead;	}	ListNode* pFirstNode  = pHead;	ListNode* pSecondNode = pHead;	while(pFirstNode->m_pNext!= NULL && pFirstNode->m_pNext->m_pNext != NULL)//如果链表结点数为偶数，输出中间两个结点前面的一个。	{		pFirstNode  = pFirstNode->m_pNext->m_pNext;		pSecondNode = pSecondNode->m_pNext;	}	return pSecondNode;}

完整测试代码：

// FindMidNodeList.cpp : 定义控制台应用程序的入口点。//#include "stdafx.h"#include <iostream>using namespace std;struct ListNode {	int  m_nValue;	ListNode* m_pNext;	ListNode(int k): m_nValue(k),m_pNext(NULL)	{	}};ListNode* FindMidNode(ListNode* pHead){	if (pHead == NULL)	{		return NULL;	}	if (pHead->m_pNext == NULL || pHead->m_pNext->m_pNext == NULL)	{		return pHead;	}	ListNode* pFirstNode  = pHead;	ListNode* pSecondNode = pHead;	while(pFirstNode->m_pNext!= NULL && pFirstNode->m_pNext->m_pNext != NULL)//如果链表结点数为偶数，输出中间两个结点前面的一个。	{		pFirstNode  = pFirstNode->m_pNext->m_pNext;		pSecondNode = pSecondNode->m_pNext;	}	return pSecondNode;}int _tmain(int argc, _TCHAR* argv[]){	ListNode* head  = new ListNode(1);	ListNode* Node1 = new ListNode(2);	ListNode* Node2 = new ListNode(3);	ListNode* Node3 = new ListNode(4);	ListNode* Node4 = new ListNode(5);	ListNode* Node5 = new ListNode(6);	ListNode* Node6 = new ListNode(7);	head->m_pNext = Node1;	Node1->m_pNext = Node2;	Node2->m_pNext = Node3;	Node3->m_pNext = Node4;	Node4->m_pNext = Node5;	Node5->m_pNext = NULL;	//Node6->m_pNext = NULL;	ListNode* p = FindMidNode(head);	cout<<p->m_nValue<<endl;	getchar();	return 0;}

C++ 算法之 查找链表中间节点