从链接列表的最后一个获取n

Question

我正在做的是，首先我扭转了链接列表，然后实际上我正在尝试获取节点的n个值。问题在于该功能在逆转链接列表后没有做任何事情，并且由于某种原因不会出现错误。

这是我的代码：

#include<stdlib.h>
#include<assert.h>

// 1. Create a linked list first
struct Node {
  int data;
  struct Node* next;
};

// 2. Create traversal function for linked list
void linkedListTraversal(struct Node* ptr) {
  while (ptr != NULL) {
    printf("%d\n", ptr->data);
    ptr = ptr->next;
  }
}

// 3. Write a function to get the node value from the tail of the linked list
int getNode(struct Node* head, int positionFromTail) {
  int value;
  struct Node* prevNode = NULL;
  struct Node* currNode = head;
  struct Node* nextNode;

  while (currNode != NULL) {
    nextNode = currNode->next;
    currNode->next = prevNode;
    prevNode = currNode;
    currNode = nextNode;
  }
  head = prevNode;

  struct Node* ptr = head;
  int count = 0;

  while (ptr != NULL) {
    if (count == positionFromTail) {
      return (ptr->data);
      count = count + 1;
      ptr = ptr->next;
    }

  }
  assert(0);
}

int main() {
  struct Node* head;
  struct Node* second;
  struct Node* third;
  struct Node* fourth;

  head = (struct Node*)malloc(sizeof(struct Node));
  second = (struct Node*)malloc(sizeof(struct Node));
  third = (struct Node*)malloc(sizeof(struct Node));
  fourth = (struct Node*)malloc(sizeof(struct Node));

  head->data = 3;
  head->next = second;

  second->data = 2;
  second->next = third;

  third->data = 1;
  third->next = fourth;

  fourth->data = 0;
  fourth->next = NULL;

  linkedListTraversal(head);

  printf("The value of the node is %d", getNode(head, 2));
}

这是我的输出，任何帮助将不胜感激。

3
2
1
0

Answer 1

您可以具有无限循环，因为仅当循环中的if语句条件评估以true

while (ptr!=NULL)
{
 if (count == positionFromTail)
 {
  return (ptr->data);
  count = count + 1;
  ptr = ptr->next;
 }


}

重写for for for for for for for ptr 至少例如，

while ( ptr != NULL && positionFromTail-- )
{
    ptr = ptr->next;
}

if ( ptr != NULL )
{
    return ptr->data;
}
else
{
    // return some value
    return  -1;
}

例如参数position> positionfromtail < /代码>应具有未签名的整数类型。否则，您需要检查函数的开头是否具有负值。

退出功能后，请注意您的列表将被打破。调用函数后，MAIN中的指针头将不会更改，但是将更改指针和其他节点指向的数据成员的值。因此，通常您的方法是不正确的。

无需扭转列表即可从列表末端计算的给定位置找到一个元素。

我会以以下方式声明函数

int getNode( const struct Node *head, int pos, int *data );

对于初学者，如果存在具有指定位置的节点或0否则， ，该函数返回1。如果有一个带有指定位置的节点，则存储的值写在“删除参数” data中。

如果参数的值pos不是负面的，则节点的计数是从列表末尾开始的。

这是一个演示程序。

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

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

void clear( struct Node **head )
{
    while (*head)
    {
        struct Node *current = *head;
        *head = ( *head )->next;
        free( current );
    }
}

size_t create( struct Node **head, const int a[], size_t n )
{
    clear( head );

    size_t i = 0;

    while (n-- && ( *head = malloc( sizeof( struct Node ) ) ) != NULL )
    {
        ( *head )->data = *a++;
        ( *head )->next = NULL;
        ++i;
        head = &( *head )->next;
    }

    return i;
}

FILE * display( const struct Node *head, FILE *fp )
{
    for (; head != NULL; head = head->next)
    {
        fprintf( fp, "%d -> ", head->data );
    }

    fputs( "null", fp );

    return fp;
}    

int getNode( const struct Node *head, int pos, int *data )
{
    int success = 0;

    if (!( pos < 0 ))
    {
        while (head != NULL && pos--)
        {
            head = head->next;
        }

        if (( success = head != NULL )) *data = head->data;
    }
    else
    {
        const struct Node *current = head;

        for ( ;current != NULL && pos; ++pos )
        {
            current = current->next;
        }

        while (current != NULL )
        {
            head = head->next;
            current = current->next;
        }

        if (( success = pos == 0 )) *data = head->data;
    }

    return success;
}

int main( void )
{
    struct Node *head = NULL;
    const int a[] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };

    create( &head, a, sizeof( a ) / sizeof( *a ) );

    fputc( '\n', display(head, stdout ) );

    for (int i = 0, data; getNode( head, i, &data ); i++)
    {
        printf( "%d: %d; ", i, data );
    }
    putchar( '\n' );

    for (int i = -1, data; getNode( head, i, &data ); i--)
    {
        printf( "%d: %d; ", i, data );
    }
    putchar( '\n' );

    clear( &head );
}

程序输出是

0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null

0: 0; 1: 1; 2: 2; 3: 3; 4: 4; 5: 5; 6: 6; 7: 7; 8: 8; 9: 9;
-1: 9; -2: 8; -3: 7; -4: 6; -5: 5; -6: 4; -7: 3; -8: 2; -9: 1; -10: 0;