我一辈子都弄不明白，我花了好几天做这个练习，但无济于事。
我正在尝试从单赞列表中删除值为0的节点。
假设我有|1| 3| 0| 4| 0| 5| 0| 0|。结果应该是|1| 3| 4| 5|
下面是所有的代码供参考

#include <iostream>
#include <fstream>

using namespace std;

struct node {
    int data;
    node* next;
};
node* head, *last;
int n;
void creating_list()
{
    node* aux;
    ifstream f("in.txt");
    f >> n;
    for(int i=0;i<n;i++)
    {
        if (head == NULL)
        {
            head = new node;
            f >> head->data;
            head->next = NULL;
            last = head;
        }
        else
        {
            aux = new node;
            f >> aux->data;
            last->next = aux;
            aux->next = NULL;
            last = aux;
            
        }
    }
    
}

void displaying_list() 
{
    node* a;
    a = head;
    if (a == NULL)
        cout << "List is empty! ";
    else
    {
        cout << "Elements of list are: | ";
        while (a)
        {
            cout << a->data<<" | ";
            a = a->next;
        }
    }
}

void delete_first_node()
{
    if (head == NULL)
        cout << "List is empty";
    else
    {
        cout << "Deleting first node\n";
        node* aux;
        aux = head;
        head = head->next;
        delete aux;
    }
}
void delete_last_node()
{
    if (head == NULL)
        cout << "List is empty";
    else
    {
        if (head == last)
        {
            delete head;
            head = last = NULL;

        }
        else
        {
            node* current;
            current = head;
            while (current->next != last)
                current = current->next;
            delete current->next;
            current->next = NULL;
            last = current;
        }
    }
}
void delete_value_0()
{
    node* aux;
    if (head == NULL)
        cout << "List is empty. Can't delete! ";
    else
    
    //  if (head->data == 0)
    //      delete_first_node();
    //  if (last->data == 0)
    //      delete_last_node();
    //  else
        {
            node* a;
            a = head;
            while (a)
                if (a->next->data != 0)
                { 
                    a = a->next;
                    cout << a->data<<" | ";
                }
                else
                    if (a->next != last)
                    {
                        aux = a->next;
                        a->next = a->next->next;
                        delete aux;
                        break;
                    }
        }

    
}

int main()
{
    creating_list();
    displaying_list(); cout <<endl;
    delete_value_0();
    
    return 0;
}

这是金属问题，我试着移动一个节点，远离值为0的节点，把值存储在另一个节点，这里是aux，然后删除aux;
我在这些行上加了注解，因为如果我不加注解，并且满足了条件，它就不会执行其余的代码...
如果我把break放在末尾，它只显示前几个数字，直到0，然后在短时间内停止，不移动整个列表。
如果我不输入break，程序就不会停止，它处于无限循环中，它不会以代码0退出

void delete_value_0()
{
    node* aux;
    if (head == NULL)
        cout << "List is empty. Can't delete! ";
    else
    
    //  if (head->data == 0)
    //      delete_first_node();
    //  if (last->data == 0)
    //      delete_last_node();
    //  else
        {
            node* a;
            a = head;
            while (a)
                if (a->next->data != 0)
                { 
                    a = a->next;
                    cout << a->data<<" | ";
                }
                else
                    if (a->next != last)
                    {
                        aux = a->next;
                        a->next = a->next->next;
                        delete aux;
                        break;
                    }
        }

    
}

老实说，我很困惑，我花了这么多时间试图弄清楚这一点，这应该是一个非常简单的练习。我觉得像回答真的很简单，但我不知道该怎么做了，也许这是不适合我。