原来错误的代码:
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SllNode *SllCopy( SllNode *const pTarget ) //拷贝单链表,若拷贝失败或参数为空链表,返回NULL,否则返回新链表的首指针;函数并不破坏原链表
{
if( NULL == pTarget )
return NULL;
SllNode *pCopy = new SllNode;
if( NULL == pCopy )
return NULL;
register SllNode *pTmpTar = pTarget;
register SllNode *pTmpCopy = pCopy;
while( NULL != pTmpTar )
{
pTmpCopy->Data = pTmpTar->Data;
pTmpCopy->pNext = pTmpTar->pNext;
pTmpTar = pTmpTar->pNext;
pTmpCopy = pTmpCopy->pNext;
}
return pCopy;
}
SllNode *SllCopy( SllNode *const pTarget ) //拷贝单链表,若拷贝失败或参数为空链表,返回NULL,否则返回新链表的首指针;函数并不破坏原链表
{
if( NULL == pTarget )
return NULL;
SllNode *pCopy = new SllNode;
if( NULL == pCopy )
return NULL;
register SllNode *pTmpTar = pTarget;
register SllNode *pTmpCopy = pCopy;
while( NULL != pTmpTar )
{
pTmpCopy->Data = pTmpTar->Data;
pTmpCopy->pNext = pTmpTar->pNext;
pTmpTar = pTmpTar->pNext;
pTmpCopy = pTmpCopy->pNext;
}
return pCopy;
}
由于才起步,经验不足,导致以上错误。
用VS2008逐条跟踪前期,并未发现此函数有错。但发现其它函数在修改SllCopy函数返回的“复制”的单链表后,原模板链表也被修改,于是想到可能是指针无意间被修改,故将参数改为const并再次调试,问题依旧。
但排除了首指针被修改问题之后,依次隔离其余各函数,发现只要调用了SllCopy函数便会出现此问题,因此将调试重点放在了此函数。
逐条跟踪局部变量,发现除模板链表首指针pTarget 和 目标链表首指针pCopy 不一样之外,其余指针都相同,便发现了问题所在。我只是简单的将原链表的各节点的pNext指针复制了一遍到pCopy所指向的目标链表,因此,函数调用后,相当于创建了一个新首指针pCopy去指向模板链表,那么通过对“复制”后的链表进行操作也就修改了原模板链表,导致后期错误的产生。
现将函数修改如下:
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SllNode *SllCopy( SllNode *const pTarget ) //拷贝单链表,若拷贝失败或参数为空链表,返回NULL,否则返回新链表的首指针;函数并不破坏原链表
{
if( NULL == pTarget )
return NULL;
SllNode *pCopy = new SllNode;
if( NULL == pCopy )
return NULL;
register SllNode *pTmpTar = pTarget;
register SllNode *pTmpCopy = pCopy;
while( NULL != pTmpTar )
{
pTmpCopy->Data = pTmpTar->Data;
pTmpCopy->pNext = new SllNode;
if( NULL == pTmpCopy->pNext )
return NULL;
pTmpCopy->pNext->pNext = NULL;
pTmpCopy = pTmpCopy->pNext;
pTmpTar = pTmpTar->pNext;
}
//Delete the useless rear node
pTmpCopy = pCopy;
while( NULL != pTmpCopy->pNext->pNext )
pTmpCopy = pTmpCopy->pNext;
pTmpCopy->pNext = NULL;
delete pTmpCopy->pNext;
return pCopy;
}
SllNode *SllCopy( SllNode *const pTarget ) //拷贝单链表,若拷贝失败或参数为空链表,返回NULL,否则返回新链表的首指针;函数并不破坏原链表
{
if( NULL == pTarget )
return NULL;
SllNode *pCopy = new SllNode;
if( NULL == pCopy )
return NULL;
register SllNode *pTmpTar = pTarget;
register SllNode *pTmpCopy = pCopy;
while( NULL != pTmpTar )
{
pTmpCopy->Data = pTmpTar->Data;
pTmpCopy->pNext = new SllNode;
if( NULL == pTmpCopy->pNext )
return NULL;
pTmpCopy->pNext->pNext = NULL;
pTmpCopy = pTmpCopy->pNext;
pTmpTar = pTmpTar->pNext;
}
//Delete the useless rear node
pTmpCopy = pCopy;
while( NULL != pTmpCopy->pNext->pNext )
pTmpCopy = pTmpCopy->pNext;
pTmpCopy->pNext = NULL;
delete pTmpCopy->pNext;
return pCopy;
}
最后删除了一个无用尾节点,但是始终觉得这几行代码应该可以不用,需要对链表复制进行怎样的改动呢?希望能得听听大家的意见。
由于才起步,代码一般很幼稚,望高手包含,更希望热心人多多指点,非常感谢。
【注】下附没有无用尾节点的SllCopy版本
SllNode *SllCopy( SllNode *const pTarget ) //拷贝单链表,若拷贝失败或参数为空链表,返回NULL,否则返回新链表的首指针;函数并不破坏原链表
{
if( NULL == pTarget )
return NULL;
SllNode *pCopy = new SllNode;
if( NULL == pCopy )
return NULL;
register SllNode *pTmpTar = pTarget;
register SllNode *pTmpCopy = pCopy;
while( NULL != pTmpTar )
{
SllNode *pTmp = new SllNode;
if( NULL == pTmp )
return NULL;
pTmp->Data = pTmpTar->Data;
pTmp->pNext = NULL;
pTmpCopy->pNext = pTmp;
pTmpCopy = pTmpCopy->pNext;
pTmpTar = pTmpTar->pNext;
}
//Delete the useless rear node
//pTmpCopy = pCopy;
//while( NULL != pTmpCopy->pNext->pNext )
// pTmpCopy = pTmpCopy->pNext;
//
//pTmpCopy->pNext = NULL;
//delete pTmpCopy->pNext;
return pCopy->pNext;
}
{
if( NULL == pTarget )
return NULL;
SllNode *pCopy = new SllNode;
if( NULL == pCopy )
return NULL;
register SllNode *pTmpTar = pTarget;
register SllNode *pTmpCopy = pCopy;
while( NULL != pTmpTar )
{
SllNode *pTmp = new SllNode;
if( NULL == pTmp )
return NULL;
pTmp->Data = pTmpTar->Data;
pTmp->pNext = NULL;
pTmpCopy->pNext = pTmp;
pTmpCopy = pTmpCopy->pNext;
pTmpTar = pTmpTar->pNext;
}
//Delete the useless rear node
//pTmpCopy = pCopy;
//while( NULL != pTmpCopy->pNext->pNext )
// pTmpCopy = pTmpCopy->pNext;
//
//pTmpCopy->pNext = NULL;
//delete pTmpCopy->pNext;
return pCopy->pNext;
}
本文转自Bill_Hoo 51CTO博客,原文链接:http://blog.51cto.com/billhoo/411476
,如需转载请自行联系原作者