下面的函数reverse() 应该反转一个单链表。函数末尾缺少一行。
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* head_ref is a double pointer which points to head (or start) pointer
of linked list */
static void reverse(struct node** head_ref)
{
struct node* prev = NULL;
struct node* current = *head_ref;
struct node* next;
while (current != NULL)
{
next = current->next;
current->next = prev;
prev = current;
current = next;
}
/*ADD A STATEMENT HERE*/
}
应该添加什么来代替“/*ADD A STATEMENT HERE*/”，以便函数正确反转链表。
(A) *head_ref = prev;
(B) *head_ref = 当前；
(C) *head_ref = 下一个；
(D) *head_ref = NULL;答案：(一)
解释： *head_ref = prev;
在while循环结束时， prev指针指向原始链表的最后一个节点。我们需要更改 *head_ref 以便头指针现在开始指向最后一个节点。
请参阅以下完整的运行程序。
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* Function to reverse the linked list */
static void reverse(struct node** head_ref)
{
struct node* prev = NULL;
struct node* current = *head_ref;
struct node* next;
while (current != NULL)
{
next = current->next;
current->next = prev;
prev = current;
current = next;
}
*head_ref = prev;
}
/* Function to push a node */
void push(struct node** head_ref, int new_data)
{
/* allocate node */
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* Function to print linked list */
void printList(struct node *head)
{
struct node *temp = head;
while(temp != NULL)
{
printf("%d ", temp->data);
temp = temp->next;
}
}
/* Drier program to test above function*/
int main()
{
/* Start with the empty list */
struct node* head = NULL;
push(&head, 20);
push(&head, 4);
push(&head, 15);
push(&head, 85);
printList(head);
reverse(&head);
printf("\n Reversed Linked list \n");
printList(head);
return 0;
}