C++ 双链接列表插入和显示以及搜索和删除

我正在实现一个双链接列表，我的搜索和删除函数给了我一个错误。 我是新的双链接列表，所以任何帮助将不胜感激

这是我的插入函数

void push(student s)
{

   if ( head == NULL)
   {
       node *tmp = new node;
       tmp->data = s;
       tmp->next = NULL;
       tmp->prev = NULL;
       head = tmp;
       tail = tmp;
   }
    else if ( head != NULL)
    {
        node *tmp = new node;
        tmp->data = s;
        tmp->next = head;
        head->prev = tmp;
        head = tmp;
    }



}

这是我的显示功能

void display()
{
    node *current = head;

    while (current!=NULL)
    {
        cout << current->data.name << endl;
        cout << current->data.GPA << endl;
        cout << current->data.id << endl;
        cout << current->data.university << endl;
        current = current->next;
    }
}

这是我的主要任务

int main()
{

    student me;
    student you;
    you.id = 9;
    you.GPA = 9.8;
    you.name = "peter";
    you.university = "stc";
    me.id = 20;
    me.GPA = 20.0;
    me.name = "robert";
    me.university = "utpa";
    student mee;
    mee.id = 15;
    mee.GPA = 8.9;
    mee.name = "mike";
    mee.university = "utpa";
    studentList list;
    list.push(me);
    list.push(you);
    list.push(mee);
    list.pop();
    list.removeStudent(15);
    list.display();

    return 0;
}

检查

head

是否为

NULL

，如果为空，则分配

head

。然后检查

head

是否不是

NULL

，此时它肯定不是上面指定的值，然后执行其他操作

简单的解决方案是，将两个

if

结构更改为

if

…

else if

，例如

if ( head == NULL)
{
   ...
   head = tmp;
}
else if ( head != NULL)
{
    ...
    head = tmp;
}

---

问题在于您的

push

功能。您可以大大简化它以避免条件逻辑问题：

void push(student s)
{
   node *tmp = new node;
   tmp->data = s;
   tmp->next = head; // this works even when head is NULL
   tmp->prev = NULL;

   if (head == NULL)
   {
       head = tmp;
       tail = tmp;
   }
   else
   {
       head->prev = tmp;
       head = tmp;
   }
}

---

即使这比它需要的要复杂一点——如果

if

条件

head==NULL

失败，那么

head

不能为NULL。瞧，没有必要测试

head！=空

，只需单独使用

else

。请参阅我对Zac答案的评论——如果

是多余的，则您的第二个
。不清楚
推
应该在前面还是后面推，您能澄清一下吗？
if ( head == NULL)
{
   ...
   head = tmp;
}
else if ( head != NULL)
{
    ...
    head = tmp;
}

void push(student s)
{
   node *tmp = new node;
   tmp->data = s;
   tmp->next = head; // this works even when head is NULL
   tmp->prev = NULL;

   if (head == NULL)
   {
       head = tmp;
       tail = tmp;
   }
   else
   {
       head->prev = tmp;
       head = tmp;
   }
}