c - C 中带头节点的链表

Question

The type definition of the list is in linkedList.h as usual.

 #ifndef LINKEDLIST_H
 #define LINKEDLIST_H

 typedef struct snode {
        int value;  
        struct snode *next; 
 } snodeType;

 typedef struct hnode {int count;
        snodeType *first;
        snodeType *last; 
 } sList;

 sList* create_sList(void);
 int insert_element_s(sList *L, snodeType *p, int value);
 int delete_element_s(sList *L, snodeType *p); 
 sList* merge_lists(sList *L1, sList *L2);

 #endif /* LINKEDLIST_H */

问题是:

sList * create_sList(void) creates the list and returns it to the caller. It has to allocate memory for the header node, initialize the fields in the struct hnode.

sList* create_sList(void) {

    sList *list = NULL;
    list->first  = (sList*)malloc(sizeof(snodeType));
    list->last  = (sList*)malloc(sizeof(snodeType));

/*  2nd option
    sList *list = NULL;
    node = malloc(sizeof(snodeType));
    node->next= NULL;
    list->first = node;
    list->last = node;
*/

    return list;
}

我只需要启动这个链表，有人知道怎么做吗？

Answer 1

那应该只是:

sList * create_sList(void)
{
  sList *list = malloc(sizof *list);
  if(list != NULL)
  {
    list->count = 0;
    list->first = list->last = NULL;
  }
  return list;
}

这将返回一个没有元素的列表头，即一个空列表头。