这是我的链接列表的remove()函数。怎么会更好,为什么?
void removeData(void *data, struct accList *theList)
{
if(theList->head == NULL) //nothing can be deleted
return;
else if(theList->head == theList->tail) //there is only one element in the list
{
free(theList->head);
theList->head = theList->tail = NULL;
}
else if(data == theList->head->data) //the node to be deleted is the head
{
struct accListNode *temp = theList->head;
free(theList->head);
theList->head = temp;
theList->head->next = temp->next;
}
else if(data == theList->tail->data) //the node to be deleted is the tail
{
struct accListNode *cur;
for(cur = theList->head; cur->next->next != NULL; cur = cur->next);
theList->tail = cur;
free(cur->next);
cur->next = NULL;
}
else //the node to be deleted is any other node
{
struct accListNode *cur;
for(cur = theList->head; cur != NULL; cur = cur->next)
{
if(cur->data == data) //this is the node we must delete from theList
{
struct accListNode *temp = cur->next->next;
free(cur->next);
cur->next = temp;
break;
}
}
}
}
另外,有人可以给我一个free()函数的详细解释。 “释放ptr指向的内存”这些词语没有帮助。
由于
答案 0 :(得分:1)
不是测试所有不同的特殊情况,而是可以使用指向列表元素指针的指针,因为无论如何你都要遍历列表,跟踪最后看到的元素:
void removeData ( void *data , struct accList *theList ) {
struct acclist *last = NULL, **finger = &theList->head;
while ( *finger != NULL ) {
if ( (*finger)->data == data )
*finger = (*finger)->next;
else {
last = *finger;
finger = &( (*finger)->next );
}
}
theList->last = last;
}
此代码与您的函数的不同之处在于它删除了匹配data
的所有元素,但您可以轻松修改它以删除第一个元素以匹配data
。