C语言 删除链接排序列表中的第一个元素

Question

我是新来的，这实际上是我的第一个问题。

我写了这段代码，我无法删除列表的第一个元素，不知何故我可以删除其他元素。 我对函数reverse_list也有问题，我相信这是因为我没有将参数作为参考传递。

我可以在 C 编程的参数中使用 & 吗？

#include <stdio.h>
#include <stdlib.h>
struct Node {
int data;
struct Node * next;
};

struct Node * build_sorted_list () {
struct Node * head = NULL, * temp, * rear, * front;
int num;
printf ("please enter a number:");
scanf ("%d", &num);
while (num != 0) {
temp = (struct Node *) malloc (sizeof (struct Node));
if (temp == NULL) {
printf ("god damn");
break;
}
temp -> data = num;
if (head == NULL) {
temp -> next = NULL;
head = temp;
}
else if (num <= head -> data) {
temp -> next = head;
head = temp;
}
else {
rear = head;
front = rear -> next;
while (front != NULL && front -> data <= num) {
rear = front;
front = front -> next;
}
temp -> next = front;
rear -> next = temp;
}
printf ("enter number please:n");
scanf ("%d", &num);
}
return head;
}

struct Node * reverse_list (struct Node * head) {
struct Node * rear, * mid, * front;
if (head != NULL || head -> next == NULL) {
return head;
}
rear = head;
mid = rear ->next;
front = mid -> next;
while (front != NULL) {
mid -> next = rear;
rear = mid;
mid = front;
front = front ->next;
}
mid -> next = rear;
head -> next = NULL;
head = mid;
return head;
}

struct Node * delete_num (int wanted, struct Node * head) {
struct Node * rear, * front;
if (head == NULL) {
return head;
}
if (head -> data == wanted) {
struct Node * temp = head;
head = head -> next;
temp = NULL;
return head;
}
rear = head;
front = head -> next;
while (front != NULL) {
if (front -> data == wanted) {
break;
}
rear = front;
front = front -> next;
}
if (front != NULL) {
rear -> next = front -> next;
}
free (front);
return head;
}

int main() {
struct Node * head;
int wanted;
head = build_sorted_list(); /* Please pretend this function exists */
reverse_list (head);
printf ("please enter a number to delete:");
scanf ("%d", &wanted);
delete_num (wanted, head);
free_list (head);
return 0;
}

Answer 1

作为管理数据结构的递归函数的经验法则。对结构的操作应返回结构。

根据定义，链表是：

• 一个空列表。
• 具有值的节点，并将链表作为其尾部。

链表具有递归性质。你需要非常小心，递归函数非常复杂，尽管代码量很短。

在你的主函数中，你有一个头指针。假设头部保持不变是不安全的。事实上，当你删除第一个元素(头部(时，你的问题就来了。

您的删除应该是这样的：

struct Node * delete_num(int wanted, struct Node * head){
if( head == NULL)
return NULL;    //Reached the end of the line, nothing to do.
if( head->data != wanted )
head->next = delete_num(wanted, head->next);    //Haven't reached it yet, but 
//the element must be down 
//the line, I want to update
//my next in case it's the 
//next node.
if(head->data == wanted){
struct Node * aux = head;
head = head->next;
free(aux);
}
return head;
}

这是假设您只想删除一个元素，并且您的列表不允许重复值。

从 main 调用此函数应为：

head = delete_num(wanted, head);

安德烈斯