题目：输入两个链表，找出它们第一个公共节点。链表节点定义如下：

struct ListNode

｛

        int    m_nKey;

        ListNode*   m_pNext;

｝

方法1：在第一个链表上顺序遍历每一个节点，没遍历一个节点，在第二个链表上顺序遍历每个节点。  O(n^2)

方法2：找到两个链表的长度差，先遍历长链表到短链表的长度处，然后两个链表同时遍历，没遍历依次比较两个节点指针是否相同，

注意是比较节点指针，不是节点的值！

代码：

// FindFirstCommandNode.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_nKey;	ListNode*   m_pNext;	ListNode(int i):m_nKey(i)	{	}};//获取链表长度int GetListLength(ListNode* pHead){	int nLength = 0;	ListNode* pNode = pHead;	while (pNode != NULL)	{		++nLength;		pNode = pNode->m_pNext;	}	return nLength;}ListNode* FindFirstCommandNode(ListNode* pHead1, ListNode* pHead2){	int  nLength1 = GetListLength(pHead1);	int  nLength2 = GetListLength(pHead2);	int nLengthDif = 0;//两个链表的长度差	ListNode* pListHeadLong  = NULL;//用于指向长链表	ListNode* pListHeadShort = NULL;//用于指向短链表	//根据长度判断 链表指向	if (nLength1 > nLength2)	{	    nLengthDif = nLength1 - nLength2;		pListHeadShort = pHead2;		pListHeadLong  = pHead1;	}	else	{		nLengthDif = nLength2 - nLength1;		pListHeadLong  = pHead2;		pListHeadShort = pHead1;	}	//先对长链表进行移动 移动到与短链表长度相同的位置	for (int i = 0; i < nLengthDif; i++)	{		pListHeadLong = pListHeadLong->m_pNext;	}	//寻找公共节点	while (pListHeadLong !=NULL && pListHeadShort != NULL && pListHeadLong!= pListHeadShort)	{		pListHeadLong  = pListHeadLong->m_pNext;		pListHeadShort = pListHeadShort->m_pNext;	}	//如果不为空  此时的pListHeadLong 与pListNodeShort为同一个节点，返回该节点	if (pListHeadLong != NULL)	{		return pListHeadLong;	}	else	{		return NULL;//否则返回NULL	}}int _tmain(int argc, _TCHAR* argv[]){	ListNode* head1 = new ListNode(0);	ListNode* head2 = new ListNode(1);	ListNode* node0 = new ListNode(22);	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);	ListNode* node8 = new ListNode(6);	head1->m_pNext = node1;	node1->m_pNext = node0;	node0->m_pNext = node3;	node3->m_pNext = node5;	node5->m_pNext = node6;	node6->m_pNext = NULL;	head2->m_pNext = node2;	node2->m_pNext = node4;	node4->m_pNext = node8;	node8->m_pNext = node6;	node6->m_pNext = NULL;	cout<<"链表1的长度为："<<GetListLength(head1)<<endl;	cout<<"链表2的长度为："<<GetListLength(head2)<<endl;	ListNode* CommNode = FindFirstCommandNode(head1,head2);	if (CommNode!= NULL)	{		cout<<"公共节点的值为："<<CommNode->m_nKey<<endl;	}	else	{		cout<<"没有公共节点"<<endl;	}	getchar();	return 0;}

C++算法之 找出两个链表的公共节点