交叉链表

两个单向链表，找出它们的第一个公共结点。链表的结点定义为：

struct ListNode

{

    int  m_nKey;      

    ListNode*   m_pNext;

};

#include "stdafx.h"
#include<stack>
#include<iostream>

using namespace std;

struct ListNode
{
	int  m_nKey;
	ListNode*   m_pNext;
};



ListNode*find_cross_node(ListNode*head1, ListNode*head2)
{
	ListNode*n1, *n2;
	stack<int>a1, a2;
	n1 = head1;
	n2 = head2;
	while (n1->m_pNext != NULL)
	{
		a1.push(n1->m_nKey);
		n1 = n1->m_pNext;
	}
	while (n2->m_pNext != NULL)
	{
		a2.push(n2->m_nKey);
		n2 = n2->m_pNext;
	}
	while (a1.top() == a2.top())
	{
		a1.pop();
		a2.pop();
		if (a1.empty())
			break;
		if (a2.empty())
			break;
	}

	int k;
	k = a2.size();
	n2 = head2;
	while (k)
	{
		n2 = n2->m_pNext;
		k--;
	}
	k = a1.size();
	n1 = head1;
	while (k)
	{
		n1 = n1->m_pNext;
		k--;
	}
	while (n1 != n2)
	{
		n1 = n1->m_pNext;
		n2 = n2->m_pNext;
	}
	return n1;
}



int _tmain(int argc, _TCHAR* argv[])
{
	ListNode*head1 = new ListNode;
	ListNode*head2 = new ListNode;
	ListNode*node1 = new ListNode;
	ListNode*node2 = new ListNode;
	ListNode*node3 = new ListNode;
	ListNode*node4 = new ListNode;
	ListNode*node5 = new ListNode;
	ListNode*node6 = new ListNode;
	head1->m_nKey = 0;
	head2->m_nKey = 0;
	node1->m_nKey = 0;
	node2->m_nKey = 0;
	node3->m_nKey = 0;
	node4->m_nKey = 4;
	node5->m_nKey = 5;
	node6->m_nKey = 6;

	head1->m_pNext = node1;
	node1->m_pNext = node2;
	head2->m_pNext = node3;
	node3->m_pNext = node4;
	node2->m_pNext = node4;
	node4->m_pNext = node5;
	node5->m_pNext = node6;
	node6->m_pNext = NULL;


	ListNode*cross_node = find_cross_node(head1, head2);
	if (cross_node == NULL)
		cout << "两个链表不交叉" << endl;

	system("pause");
	return 0;
}

思路是：从两个链表交叉点开始，后面的都重合，因此很自然将两个链表末尾对齐，对齐方法是从后往前一个节点一个节点的比较值，若不相等，则从这两个链表不相等位置的下一个开始一一比较就行了。

复杂度应该是O(n)。