如何删除链表中连续的两项

void delete_double (LN<T>*& l) {
    if (l == nullptr)
        return;

    LN<T> *p = l;
    while ( p -> next != nullptr && p -> next -> next != nullptr)
    {
        if (p -> value == p -> next -> value) // the current value is equal to the next value in the linked list 
        {
            if (p == l)                     // when the first two values are the same                                        // not sure if it is correct
            {
                l = l -> next -> next;
            }
            else                            // Problem should be here
            {
                LN<T> *to_delete = p;       //  Also tried this (doesn't work)
                p = p->next;
                delete to_delete;           //  LN<T>* to_delete = p;
                                            //  LN<T>* to_delete2 = p -> next;
                LN<T> *to_delete1 = p;      //  l = to_delete2 -> next;
                p = p->next;                //  delete to_delete;
                delete to_delete1;          //  delete to_delete2;
            }
        }
        else
        {
            p = p-> next;
        }
    }
}
//  Image below is my output

在此处输入图片描述

我正在编写一个函数，如果两个值相同，它将删除链表中连续的两个值.当输入类似于1 -> 2 -> 3 -> 3 -> 4 -> nullptr"时，我的代码似乎停止工作.(输出应为 1 -> 2 -> 4 -> nullptr).它退出而不给我任何错误.我一行一行地调试，它突然退出并显示变量不可用".

   void delete_double(LN<T>*& l) {

        if (l == nullptr)
            return;

        LN<T> *p = l;
        LN<T> dummy(0);
        dummy.next = l;
        p = &dummy;

        LN<T> *temp;
        LN<T> *duplicate;
        LN<T> *prev;

        while (p != nullptr && p->next != nullptr)
        {
            temp = p;
            while (p != nullptr && temp->next != nullptr)
            {
                if (p->value == temp->next->value)
                {
                    duplicate = temp->next;
                    temp->next = temp->next->next;
                    delete duplicate;

                    duplicate = p;
                    prev->next = p->next;
                    p = prev;
                    delete duplicate;

                    temp = p;
                }
                else
                {
                    break;
                }
            }
            prev = p;
            p = p->next;
        }

        l = dummy.next;
    }