gpt4 book ai didi

c - 删除链接排序列表中的第一个元素

转载 作者:行者123 更新时间:2023-11-30 16:30:55 26 4
gpt4 key购买 nike

我是新来的,这实际上是我的第一个问题。

我编写了这段代码,但无法删除列表的第一个元素,但我可以删除其他元素。我的函数reverse_list也有问题,我相信这是因为我没有将参数作为引用传递。

在 C 编程中我可以在参数中使用 & 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;
}

最佳答案

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,链接列表是:

  • 空列表。
  • 一个有值的节点,尾部有一个链表。

链表具有递归性质。你需要非常小心,尽管代码量很短,但递归函数却非常复杂。

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,当您删除第一个元素(head)时,您的问题就出现了。

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的值。

从 main 调用此函数应该是:

head = delete_num(wanted, head);

安德烈斯

关于c - 删除链接排序列表中的第一个元素,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/50842694/

26 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com