C++ 我不编译。iostream库提供对NULL的支持，我不知道nullptr。至于您第一次回答的最后一部分：我试图说有三个节点：前一个节点、当前节点和下一个节点。如果下一个节点（curr->next）不是null ptr，则将上一个节点指向该节点（prev->

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我不编译。iostream库提供对NULL的支持，我不知道nullptr。至于您第一次回答的最后一部分：我试图说有三个节点：前一个节点、当前节点和下一个节点。如果下一个节点（curr->next）不是null ptr，则将上一个节点指向该节点（prev->next=curr->next），并删除当前节点。我不确定如何将previous定义为null/nullptr。我想这是我需要解决的问题。

nullptr

是在c++11中添加的。我提到它是因为问题被标记了。如果它不起作用，那就意味着它没有被设置为编译c++11。哇，太棒了，我不知道我能做一些像“prev->next->prev=prev”的事情；事实是，你根本不需要tmp。你需要做2个赋值，prev需要跳过curr，然后，它现在指向的东西，需要指向它。或者第二个可以在if语句中，check prev->next现在为null\u ptr，这将减少对上面列表的完整单独结尾的需要。这就是我尝试使用tmp指针的原因，但不知道如何做到这一点。您需要删除中间的节点，并确保->下一个的下一个点和->下一个点。

class node {  

public:

    // default constructor
    node() {name = ""; prev = NULL; next = NULL;};

    // default overloaded
    node(string s) {name = s; prev = NULL; next = NULL;};

    // item in the list
    string name; 

    // links to prev and next node in the list
    node * next, * prev;

};

tmp->prev = prev;

void linkedList::remove(string s) 
{
        bool found = false;
        node * curr = getTop(), * prev = NULL;
        node * tmp = new node();
        while(curr != NULL) 
        {
                 // match found, delete
                 if(curr->name == s) 
                 {
                        found = true;
                        // found at top
                        if(prev == NULL) 
                        {
                            node * temp = getTop();
                            setTop(curr->next);
                            getTop()->prev = NULL;
                            delete(temp);
                        } // end if
                        else 
                        {
                            // determine if last item in the list
                            if (curr->next = NULL) 
                            {
                                // prev node points to next node
                                prev->next = curr->next;
                                // delete the current node
                                delete(curr);
                            } // end if
                            // if not last item in list, proceed as normal
                            else 
                            {
                                // prev node points to next node
                                prev->next = curr->next;
                                // set the next node to its own name
                                tmp = prev->next;
                                // set prev-link of next node to the previous node (aka node before deleted)
                                tmp->prev = prev;
                                // delete the current node
                                delete(curr);
                            } // end else
                        } // end else
                    } // end if

                // not found, advance pointers
                if(!found) 
                {
                    prev = curr;
                    curr = curr->next;
                } // end if

                // found, exit loop
                else curr = NULL;
        } // end while

        if(found)
            cout << "Deleted " << s << endl;
        else 
            cout << s << " Not Found "<< endl;
} // end remove

Node *cur = find("abcd");
Node *prev = cur->prev;
prev->next = cur->next;

Node *n = cur->next;
n->next = cur->prev;

cur->next = NULL; //or nullptr
cur->prev = NULL; //or nullptr

delete cur;

if (curr->next = NULL) { ...

if (curr->next == nullptr) { ...

prev->next = curr->next;

prev->next = curr->next;
prev->next->prev = prev;
delete (curr);

void linkedList::remove(const std::string& s)
{
    node* current = getTop(); // get head node
    while (current != nullptr) // find the item you are trying to remove
    {
        if (current->name == s)
        {
            break; // when you find it, break out of the loop
        }
    }

    if (current != nullptr) // if found, this will be non-null
    {
        if (current->prev) // if this is not the head node
        {
            current->prev->next = current->next;
        }
        else
        {
            // update head node
        }

        if (current->next) // if this is not the tail node
        {
            current->next->prev = current->prev;
        }
        else
        {
            // update tail  node
        }

        // at this point, current is completely disconnected from the list
        delete current;
    }
}