4，常数时间内删除单链表中某结点

问题：

一个单链表，很长，遍历一遍很慢，我们仅知道一个指向某节点的指针pNode，而我们又想删除这个节点。

思路：

将该结点的下一个结点的内容拷贝到当前结点，然后删除下一个结点。

如果该结点时最后一个结点，则必须遍历一遍单链表，找到前面一个结点，然后删掉该结点。

注意：

虽然最后一种情况是线性复杂度的，但是总体上来看还是常数时间的。

// LinkTable.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include <iostream>
#include <string>
using namespace std;

//链表的结构体
struct node
{
	char val;
	node * next;
};

//2，找第4个结点
struct node * create( string & str_link )
{
	int len = str_link.length();

	struct node * phead = new node();     //带有表头的链表，表头中不存储任何元素
	struct node * preNode = phead;
	for( int i=0; i<len; i++ )
	{
		struct node * pNode = new node();
		pNode->val = str_link[i];
		pNode->next = NULL;
		preNode->next = pNode;
		preNode = pNode;
	}
	return phead;
}

void out_link( struct node * phead )
{
	if( phead == NULL )
		return;
	struct node * pNode = phead->next;
	while( pNode )
	{
		cout <<pNode->val;
		pNode = pNode->next;
	}
	cout << endl;
}

struct node * find_third_node( struct node * phead )
{
	if( !phead ) return NULL;

	//快指针先走4步
	struct node *pFast = phead;
	struct node *pSlow = phead;
	int num =4;
	while(num--)
	{
		pFast = pFast->next;
		if(!pFast)    //还没走完就没了
		{
			return NULL;
			break;
		}
	}

	//快慢指针一起走
	while(pFast)
	{
		pFast = pFast->next;
		pSlow = pSlow->next;
	}

	return pSlow;
}

struct node * find_last_node( struct node * phead )
{
	struct node *pNode = phead;
	while(pNode->next)
	{
		pNode = pNode->next;
	}
	return pNode;
}

void delete_node( struct node * phead, struct node * pNode )
{
	if( !phead ) return;

	//如果pNode是最后一个结点，则找到前面一个结点
	if( !pNode->next )
	{
		struct node *preNode = phead;
		while(phead)
		{
			if(preNode->next==pNode)
				break;
			else
				preNode = preNode->next;
		}
		if( !preNode )
		{
			cout <<"The node is not in the link list" << endl;
			return;
		}
		preNode->next = NULL;
		delete pNode;
	}
	else          //否则拷贝下一个元素的内容，并删除下一个结点
	{
		struct node * pNext = pNode->next;
		pNode->val = pNext->val;
		pNode->next = pNext->next;
		delete pNext;
	}
}

void test()
{
	string str;
	cin >> str;
	struct node *phead = create( str );
	cout << "The Link is : ";
	out_link( phead );
	struct node *pNode = find_third_node(phead);
	
	delete_node( phead, pNode );
	cout << "delete the last third node:" <<endl;
	out_link( phead );
	pNode = find_last_node(phead);
	delete_node( phead, pNode );
	cout << "delete last node: "<<endl;
	out_link( phead );

}

int _tmain(int argc, _TCHAR* argv[])
{
	test();
	return 0;
}