如何在双向链表中实现擦除方法

Question

我必须编写一个双链接列表，我正在尝试实现erase（Type obj）方法，该方法接受一个参数obj，遍历列表并删除包含元素obj的每个节点。

我面临的问题是，我从前面遍历链表，当我找到包含obj元素的节点时，我会在包含obj元素的节点前后更改节点的下一个/上一个指针。然而，我并没有删除带有obj本身的节点，据我所知，c没有垃圾收集，所以带有obj的节点仍然悬在空中。我如何删除它？

我的擦除（）

template <typename Type>
int Double_list<Type>::erase( Type const &obj ) {
if (empty()){
    return 0;
}
if (size() == 1 && head()->retrieve() == obj){
    list_head = nullptr;
    list_tail = nullptr;
    list_size--;
    return 1;
}

//Counter to hold the number of items deleted
int count = 0;
    //Iterating through the linked list
for (Double_node<Type> *ptr = head(); ptr != nullptr; ptr = ptr->next()){
    if (ptr->retrieve() == obj){
        ptr->previous_node->next_node = ptr->next();
        ptr->next()->previous_node = ptr->previous();
        count++;
        // delete ptr; // This stops me from iterating through the for loop
        list_size--;
    }
}
return count;
}

Answer 1

删除指针和对象时有一种称为delete的语法。示例代码：

obj *temp = getCurrentNode();
//set pointers in nodes to the correct places
delete temp; //This releases all of the memory used in the object back to the OS
temp = NULL;//good practice to set this to null, since it points to non-allocated memory
//but exiting the function will make the pointer 'temp' useless, anyway.

Answer 2

当你遍历你的列表时，你正在使用指向节点的指针，它的类型是Double_node

Double_node<Type> *ptr_next = 0;
for (Double_node<Type> *ptr = head(); ptr != nullptr; ptr = ptr_next) {
     ptr_next = ptr->next ();
     if (ptr->retrieve() == obj){
     if (ptr->previous_node)
       ptr->previous_node->next_node = ptr->next();

     ptr->next()->previous_node = ptr->previous();
     count++;
     list_size--;
     delete ptr;
    }

我相信这应该能奏效。