c - 在列表的开头插入节点

Question

我需要在列表的开头插入一个节点，我该怎么做？

使用此代码:

while(tmp!=NULL){
    printf("__________");
    printf("\nCodigo: %d\n",tmp->code);
    printf("Nombre: %s\n",tmp->name);
    printf("Apellido: %s\n",tmp->last);
    tmp = tmp->next;

};

我打印列表，这是我看到的:

---

科迪戈:3
姓名:第三名
阿佩利多:节点

---

科迪戈:2
名称:secc
阿佩利多:节点

---

科迪戈:1
名称:第一
阿佩利多:节点

所以，如果我在开头插入一些内容，我应该会看到

---

科迪戈:3
姓名:第三名
阿佩利多:节点

---

科迪戈:2
名称:secc
阿佩利多:节点

---

科迪戈:1
名称:第一
阿佩利多:节点

---

科迪戈:4
名号:第四名
阿佩利多:节点

我该怎么做？我试过这个:

    tmp_aux = lista;// creating an aux list
    while(tmp_aux->next!=NULL){
        tmp_aux->next = tmp_aux;
    }; // i used this becouse the last printed (up) is the first node
    new_aux = (struct nodo* ) malloc(1*sizeof(struct nodo));
    printf("ingrese el codigo: ");
    scanf("%d",&(*new_aux).code);
    printf("ingrese el nombre: ");
    scanf("%s",&(*new_aux).name);
    printf("ingrese el apellido: ");
    scanf("%s",&(*new_aux).last);



    new_aux->next = tmp_aux;// then i put the aux on the next of my new node
    lista = new_aux;// and make my list the new one

Answer 1

我个人认为应该首先打印第一个节点(引用评论)，但我认为这只是语义。

在我使用链表的所有时间里，我都使用了head 和tail 指针。 head 指针指向列表中的第一项，tail 指针指向列表中的最后一项。每次从列表中添加和删除项目时，都需要一些额外的簿记来使这些保持最新，但我认为这是值得的。任何需要您遍历列表的操作(搜索特定节点、打印所有项目等)都可以更简单地完成，因为您从 head 开始并转到 tail。像下面这样的东西应该让你开始，这并不意味着是一个包罗万象的程序:

 static struct nodo *head = NULL, *tail = NULL;

 struct nodo* insert_at_head(struct nodo* new_aux)
 {
   if (head == NULL && tail == NULL)
   {
     // our list is empty; any item inserted is both the beginning and end
     head = new_aux;
     tail = new_aux;
     new_aux->next = NULL;  // only 1 item in the list, there is no next element
   }
   else
   {
     // if maintained properly, this should be the only other possibility
     new_aux->next = head;  // new_aux is the new head of the list, so the previous head is now the 2nd item
     head = new_aux;  // make new_aux the new head of the list
   }

   // in fact, since head = new_aux happens in both branches, that should just go here

   return head;  // this could be a void function, but returning head and checking that it equals new_aux shows that new_aux is now the head of the list
 }

 struct nodo* remove_head()
 {
   if (head != NULL)  // our list is not empty, so it does in fact have a head
   {
     struct nodo* temp = head
     head = head->next;  // even if there is one item in the list, head->next should be NULL, so now head is NULL
     free(temp);
   }
   else
   {
     // this means our list is empty, optionally print an error message or warning "Trying to delete head from empty list!"
     return NULL;
   }

   return head;
 }

 // now iterating over all the nodes is easy, you just have to go from head to tail.
 void print_list()
 {
   struct nodo* cur_aux;
   for (cur_aux=head; cur_aux!=NULL; cur_aux=cur_aux->next)
   {
     // print whatever you want here
   }
 }

 // you can have several other functions, for manipulating the list.  Their prototypes *might* look like the following:
 // do not forget to maintain head and tail pointers for all of these!
 struct nodo* insert_at_tail(stuct nodo* new_aux); // similar to insert_at_head(), except now you want to make the current last node the 2nd to last node
 struct nodo* locate_aux(const struct nodo* aux); // iterate head to tail and return the node that matches all fields of struct nodo, return NULL if not found
 void delete_aux(struct nodo* aux); // iterate through the list and delete aux if found
 void clean_up_list(); // iterate from head to tail and delete all items
 struct nodo* insert_aux_after(struct nodo* insert_after, struct nodo* new_aux); // this will insert new_aux after insert_after

 int main(int argc, char* argv[])
 {
   // something like this
   struct nodo* new_aux = malloc(sizeof(struct nodo));
   struct nodo* new_aux2 = malloc(sizeof(struct nodo));
   struct nodo* new_aux3 = malloc(sizeof(struct nodo));

   // fill in the fields for each new_aux
   if (insert_at_head(new_aux) != new_aux)
   {
     // some error happened on insertion,, handle it
   }
   insert_at_head(new_aux2);
   insert_at_head(new_aux3);

   print_list();
   // the output should print new_aux3, then new_aux2, and finally new_aux

   clean_up_list();
   return 0;
 }

您可以将 head 和 tail 调整为列表中的第一项或最后一项，但一般约定将 head 标记为列表中的第一项列表。我可能会为其他原型(prototype)填写一些代码。事实上，你可以在没有 tail 指针的情况下实现上面的所有内容，只需在 head 处开始对列表的所有迭代，然后一直到 ->next == NULL。您还可以考虑维护一个 static size_t num_aux 来保持列表中项目数量的运行计数。当尝试从列表中删除项目时，这对于确定成功或失败特别有帮助。我怀疑如果你用 google 搜索链表教程，你会得到比我提供的代码好得多的代码，但我展示的至少应该是处理链表的一种合理方法。