如何计算链表中的节点数？为什么输出显示节点数为“2”？

Question

我编写了一个 c 程序来查找 link 列表中的节点数。但是问题出现了，因为我正在打印的计数值是“2”。

我的确切输出看起来像->

节点数为 2

我做错了什么？

//the code for the program is here:-

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

struct node
{
 int data;
 struct node *next;
}*first=NULL;

void create(int a[],int n)
{
 struct node *t,*last;
 first=(struct node *)malloc(sizeof(struct node));
 first->data=a[0];
 first->next=0;
 last=first;
 int i;
 for(i=1;i<n;i++)
 {
  t=(struct node *)malloc(sizeof(struct node));
  t->data=a[i];
  t->next=NULL;
  last->next=t;
  last=first;
 }
}

void count(struct node *p) //function to count number of nodes
{
 int count=0;
 while(p!=NULL)
 {
  count++;
  p=p->next;
 }
 printf("Number of nodes are %d ",count);
}

int main()
{
 int a[]={1,2,3,4};
 create(a,4);
 count(first);
 return 0;
}

Answer 1

我认为在函数内循环create

for(i=1;i<n;i++)

你是说

last = last->next;

而不是

last=first;

注意，当函数依赖于全局变量时，将指向头节点的指针声明为全局指针是一个坏主意。

此外，用户可以将数组中等于 0 的元素的数量传递给函数。内存分配也可能会失败。

我将按以下方式声明函数

size_t create( struct node **head, const int a[], size_t n )
{
    // if the list is not empty free its nodes
    while ( *head != NULL )
    {
        struct node *current = *head;
        *head = ( *head )->next;
        free( current );
    }

    size_t i = 0;

    for ( ; i < n && ( *head = malloc( sizeof( struct node ) ) ) != NULL; i++ )
    {
        ( *head )->data = a[i];
        ( *head )->next = NULL;

        head = &( *head )->next;
    }

    return i;
}

并像这样调用函数

size_t n = create( &first, a, sizeof( a ) / sizeof( *a ) );

在这种情况下，函数 returns 列表中创建的节点数。

这是一个演示程序。

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

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

size_t create( struct node **head, const int a[], size_t n )
{
    // if the list is not empty free its nodes
    while ( *head != NULL )
    {
        struct node *current = *head;
        *head = ( *head )->next;
        free( current );
    }

    size_t i = 0;

    for ( ; i < n && ( *head = malloc( sizeof( struct node ) ) ) != NULL; i++ )
    {
        ( *head )->data = a[i];
        ( *head )->next = NULL;

        head = &( *head )->next;
    }

    return i;
}

void output( const struct node *head )
{
    for ( ; head != NULL; head = head->next )
    {
        printf( "%d -> ", head->data );
    }

    puts( "null" );
}

int main(void) 
{
    struct node *head = NULL;

    int a[] = { 1, 2, 3, 4 };
    const size_t N = sizeof( a ) / sizeof( *a );

    size_t n = create( &head, a, N );

    printf( "There are %zu nodes in the list\n", n );

    printf( "They are " );

    output( head );

    return 0;
}

它的输出是

There are 4 nodes in the list
They are 1 -> 2 -> 3 -> 4 -> null

如何计算链表中的节点数？为什么输出显示节点数为“2”？

How to count number of nodes in a linked-list ?Why does the output show the count of nodes as '2'?

c

struct

linked-list

data-structures

singly-linked-list