C 按用户选择旋转链表

如何解决C 按用户选择旋转链表

我编写了一个程序，它从用户编号到链表和编号，以了解每个节点向左旋转多少。我只是成功地做到了这一点，但不是在一个圈子里。并且我的程序需要能够将节点向左移动，然后再将列表的长度移动到圆圈中。 有人知道我该如何修复我的程序？？ （需要修复的函数是“RotateALinkedList”函数）。我的意思是如果用户想要将列表向左移动 4 次，第一个节点将从最后一个节点开始。

#include<stdio.h>
#include<stdlib.h>


typedef struct numbers_list
{
    int data;
    struct numbers_list* next;
}number;

void RotateALinkedList(number** head,int node); //the function that rotate the linked list
int CreateLinkedList(number** head,int iNumberofNode);
int attachToEnd(number** head,int k);
void PrintTheList(number* pNode);
void FreeAllocatedMemory(number** head);

int main(void)
{
    int list_len = 0;
    int data = 0;
    number* head = NULL;
    printf("How many nodes in list? ");
    scanf("%d",&list_len);
    getchar();
    CreateLinkedList(&head,list_len);
    printf("Choose a number k,and the list will be rotated k places to the left: ");
    scanf("%d",&data);
    getchar();
    if (data <= list_len)
    {
        RotateALinkedList(&head,data);
        PrintTheList(head);
    }
    else
    {
        printf("Please Enter Valid number of node\n");
    }
    FreeAllocatedMemory(&head);
    getchar();
    return 0;
}

void RotateALinkedList(number** head,int node)
{
    int count = 0;
    number* p = *head;
    number* tempNode = NULL;
    for (count = 1; ((count < node) && (p != NULL)); count++)
    {
        p = p->next;
    }
    if (p == NULL)
    {
        return;
    }
    else
    {
        tempNode = p;
    }
    while (p->next != NULL)
    {
        p = p->next;
    }
    p->next = *head;
    *head = tempNode->next;
    tempNode->next = NULL;
}

int CreateLinkedList(number** head,int iNumberofNode)
{
    int data = 0;
    int iRetValue = -1;
    int count = 0;
    number* pNewNode = NULL;
    for (count = 0; count < iNumberofNode; count++)
    {
        printf("Enter number: ");
        scanf("%d",&data);
        getchar();
        if ((*head) == NULL)
        {
            pNewNode = (number*)malloc(sizeof(number));
            if (pNewNode != NULL)
            {
                pNewNode->data = data;
                pNewNode->next = NULL;
                *head = pNewNode;
                iRetValue = 0;
            }
        }
        else
        {
            iRetValue = attachToEnd(head,data);
        }
    }
    return iRetValue;
}

int attachToEnd(number** head,int k)
{
    int iRetValue = -1;
    number* pLastNode = NULL;
    number* pNewNode = NULL;
    pLastNode = *head;
    pNewNode = (number*)malloc(sizeof(number));
    if (pNewNode != NULL)
    {
        pNewNode->data = k;
        pNewNode->next = NULL;
        iRetValue = 0;
    }
    if (pLastNode == NULL)
    {
        *head = pNewNode;
    }
    else
    {
        while (pLastNode->next != NULL)
        {
            pLastNode = pLastNode->next;
        }
        pLastNode->next = pNewNode;
    }
    return iRetValue;
}

void PrintTheList(number* pNode)
{
    printf("the rotated list:\n");
    while (pNode != NULL)
    {
        printf("%d  ",pNode->data);
        pNode = pNode->next;
    }
}

void FreeAllocatedMemory(number** head)
{
    number* ptempNode = NULL;
    number* pFirstNode = NULL;
    pFirstNode = *head;
    while (pFirstNode != NULL)
    {
        ptempNode = pFirstNode;
        pFirstNode = pFirstNode->next;
        free(ptempNode);
    }
    *head = NULL;
}

解决方法

对于初学者来说，这个 typedef 声明中别名的名称 number

typedef struct numbers_list
{
    int data;
    struct numbers_list* next;
}number;

令人困惑。

最好定义两个结构体

struct Node
{
    int data;
    struct Node *next;
};

struct List
{
    size_t size;
    struct Node *head;
};

并在主列表中定义

struct List numbers = { .size = 0,.head = NULL };

如果列表包含指定列表中节点数的数据成员，您将简化您的任务。

函数 CreateLinkedList 应该完成一项任务：从用户指定的数组创建一个列表。它应该要求用户输入将存储在列表中的数字。

我可以建议以下演示程序中显示的以下解决方案。

#include <stdio.h>
#include <stdlib.h>

struct Node
{
    int data;
    struct Node *next;
};

struct List
{
    size_t size;
    struct Node *head;
};

void clear( struct List *list )
{
    while ( list->head )
    {
        struct Node *tmp = list->head;
        list->head = list->head->next;
        free( tmp );
    }
    
    list->size = 0;
}

size_t create( struct List *list,const int a[],size_t n )
{
    clear( list );
    
    for ( struct Node **current = &list->head; 
          n-- && ( *current = malloc( sizeof( struct Node ) ) ) != NULL; )
    {
        ( *current )->data = *a++;
        ( *current )->next = NULL;
        current = &( *current )->next;
        ++list->size;
    }
    
    return list->size;
}

FILE * display( const struct List *list,FILE *fp )
{
    fprintf( fp,"There is/are %zu items: ",list->size );
    for ( struct Node *current = list->head; current != NULL; current = current->next )
    {
        fprintf( fp,"%d -> ",current->data );
    }
    
    fputs( "null",fp );
    
    return fp;
}

void rotate( struct List *list,size_t n )
{
    if ( ( list->size != 0 ) &&  ( ( n %= list->size ) != 0 ) )
    {
        struct Node **current = &list->head;
        
        while ( n-- ) current = &( *current )->next;
        
        struct Node *tmp = list->head;
        
        list->head = *current;
        *current = NULL;
        
        struct Node *last = list->head;
        while ( last->next != NULL ) last = last->next;
        last->next = tmp;
    }
}

int main(void) 
{
    struct List numbers = { .size = 0,.head = NULL };
    int a[] = { 0,1,2,3,4,5,6,7,8,9 };
    
    create( &numbers,a,sizeof( a ) / sizeof( *a ) );
    
    fputc( '\n',display( &numbers,stdout ) );
    
    rotate( &numbers,1 );
    
    fputc( '\n',2 );
    
    fputc( '\n',3 );
    
    fputc( '\n',4 );
    
    fputc( '\n',stdout ) );
    
    clear( &numbers );
    
    return 0;
}

程序输出为

There is/are 10 items: 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null
There is/are 10 items: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 0 -> null
There is/are 10 items: 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 0 -> 1 -> 2 -> null
There is/are 10 items: 6 -> 7 -> 8 -> 9 -> 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> null
There is/are 10 items: 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null