这是在双向链表中的给定位置插入数据的相当好的方法吗?

Is this considerably good way to insert data at given position in doubly Linked List?

这个 InsertAt() 函数可以用更好的方式完成吗?

我需要很好的计算机编程指南,可以帮助我提高我的编码技能 1对1方式。我写这行是因为我的问题因为太少而没有提交 细节,因为我无话可说。 这在 Whosebug 中是个小麻烦

代码

#include <stdio.h>
#include <stdlib.h>
struct Node *Head;
struct Node
{
    char data;
    struct Node *prev;
    struct Node *next;
};
//this function inserts data at head.
void Insert(char data)
{
    struct Node *tempHead = (struct Node *)malloc(sizeof(struct Node));
    tempHead->data = data;
    tempHead->next = Head;
    tempHead->prev = NULL;
    if (Head != NULL)
        Head->prev = tempHead;
    Head = tempHead;
}
void Print()
{
    struct Node *tempHead = Head;
    while (tempHead != NULL)
    {
        printf("%c,", tempHead->data);
        tempHead = tempHead->next;
    }
    printf("\n");
}
// this function inserts at given pos .
void InsertAt(int pos)
{
    struct Node *tempHead = Head;
    struct Node *container = (struct Node *)malloc(sizeof(struct Node));
    container->data = 'k';
    if (pos == 1)
    {
        Head->prev = container;
        container->prev = NULL;
        container->next = Head;
        Head = container;
        return;
    }
    else
    {
        for (int i = 1; i < pos; i++)
        {
            if (tempHead->next == NULL && i != (pos - 1))
            {
                printf("Data OverIndexed at  %d .\n",pos);
                break;
            }
            else if (tempHead->next == NULL && i == pos - 1)
            {
                tempHead->next = container;
                container->prev = tempHead;
                container->next = NULL;
                break;
            }

            else if (i == pos - 1)
            {
                struct Node *temp1 = tempHead->next;
                tempHead->next = container;
                container->prev = tempHead;
                container->next = temp1;
                temp1->prev = container;
                break;
            }

            tempHead = tempHead->next;
        }
    }
  
}

int main()
{
    Head = NULL;
    Insert('a');
    Insert('b');
    Insert('c');
    Insert('d');
    Insert('e');
    Insert('f');
      Print();
    InsertAt(1);
      Print();
    InsertAt(5);
      Print();
    InsertAt(12);
      Print();
    InsertAt(2);
      Print();
    InsertAt(20);
}

输出

f,e,d,c,b,a,
k,f,e,d,c,b,a,
k,f,e,d,k,c,b,a,
Data OverIndexed at  12 .
k,f,e,d,k,c,b,a,
k,k,f,e,d,k,c,b,a,
Data OverIndexed at  20 .

Can this InsertAt() function thing be done in more better way?

是的,可以。您的代码中有几个错误。

试试这个:

int main()
{
    Head = NULL;
    InsertAt(1);
    Print();
}

它崩溃了吗?

您在此处取消引用 Head 而不检查 NULL:

if (pos == 1)
{
    Head->prev = container;  <--- what if Head is NULL ?

另一个错误:

如果函数 InsertAtpos<= 0 时被调用,则函数静默 returns,即当 pos 时没有像 done 那样打印错误太高了。

还有一个错误:

在无法插入新元素的情况下,即 pos <= 0pos > length-of-list,您会泄漏内存。您已经使用 malloc 来获取一个新节点,但是当插入没有发生时,内存就会“丢失”。不要malloc在知道你是否需要新节点之前。

除此之外:

  1. 使用全局变量真是个坏主意Head

  2. 可能会失败(插入新节点)的函数应该有一个 return 值来指示 succes/failure

  3. 很奇怪 InsertAt 没有取 data 值(像 Insert)。

我会做类似的事情:

int InsertAt(struct Node **pHead, int pos, char data)
{
    if (pos <= 0) return -1;   // Illegal pos

    if (pos == 1)
    {
        // pos == 1 is always valid
        struct Node *container = malloc(sizeof *container);
        if (container == NULL) exit(1);
        container->data = data;
        container->prev = NULL;
        container->next = *pHead;
        if (container->next != NULL) container->next->prev = container;
        *pHead = container;  // Change Head
        return 0;
    }

    if (*pHead == NULL) return -1;   // Illegal pos

    struct Node *tempHead = *pHead;
    for (int i = 1; i < (pos-1); i++)  // Notice pos-1
    {
        tempHead = tempHead->next;
        if (tempHead == NULL) return -1;  // Illegal pos
    }

    // Insert after tempHead
    struct Node *container = malloc(sizeof *container);
    if (container == NULL) exit(1);
    container->data = data;
    container->prev = tempHead;
    container->next = tempHead->next;
    tempHead->next = container;
    if (container->next != NULL) container->next->prev = container;

    return 0;
}

并像这样打电话:

struct Node *Head = NULL;
if (InsertAt(&Head, 1, 'b')) puts("InsertAt failed");
if (InsertAt(&Head, 2, 'd')) puts("InsertAt failed");
if (InsertAt(&Head, 2, 'c')) puts("InsertAt failed");
if (InsertAt(&Head, 1, 'a')) puts("InsertAt failed");