c-从双链接列表中删除给定字符串出现次数相同的所有节点

Question

Am正在尝试删除给定字符串的所有出现。此代码删除一个节点，但保留所有其他节点。我试着把它绕一圈，但仍然没有效果。

我还想知道，是否有可能删除结构参数并使其发生。我试图删除它并修改代码，但我得到了一个分段错误。感谢您提前回答。

void deleteNode(struct node** head, char* searchName) {
struct node* current;
for(current=*head; current; current = current->next){
if (!strcmp(current->name, searchName))
{
if (current->next)
current->next->prev = current->next;
if (current->prev)
current->prev->next = current->prev;
if (current == *head)
*head = current->next;
current->prev = NULL;
current->next = NULL;
free(current);
return;
}
}
}

Answer 1

return语句将在第一次命中后结束函数。此外，序列current->next = NULL; free(current); current = current->next将在释放内存后访问内存，并将current设置为NULL。您可能需要在循环中更新current，并使用一个额外的节点指针变量进行删除。

void deleteNode(struct node** head, char* searchName)
{
struct node* current;
struct node* to_delete;
for(current=*head; current;)
{
if (!strcmp(current->name, searchName))
{
to_delete = current;
current = current->next;
if (current) {
current->prev = to_delete->prev;
}
if (to_delete->prev) {
to_delete->prev->next = current;
}
if (to_delete == *head)
*head = current;
free(to_delete);
} else {
current = current->next;
}
}
}

Answer 2

您的函数错误。例如，如果语句

if (current->next)
current->next->prev = current->next;

应像一样重写

if (current->next)
current->next->prev = current->prev;
^^^^

功能可以通过以下方式声明和定义

void deleteNode( struct node **head, const char *searchName )
{
while ( *head )
{
if ( strcmp( ( *head )->name, searchName ) == 0 )
{
struct node *current = *head;
if ( current->next ) current->next->prev = current->prev;
*head = current->next;

free( current );
}
else
{
head = &( *head )->next;
}
}
}

请注意，如果结构struct node的数据成员name指向存储字符串的动态分配内存，则在为上述函数中的指针current调用函数free之前，必须释放为该字符串分配的内存。也就是说，您需要再添加一个类似的函数free的调用

free( current->name );
free( current );

如果你有一个指向头节点的全局指针，例如head，那么函数可以用以下方式定义

void deleteNode( const char *searchName )
{
struct node **head_ref = &head;
while ( *head_ref )
{
if ( strcmp( ( *head_ref )->name, searchName ) == 0 )
{
struct node *current = *head_ref;
if ( current->next ) current->next->prev = current->prev;
*head_ref = current->next;

free( current );
}
else
{
head_ref = &( *head_ref )->next;
}
}
}