Question

只有当节点的总和与列表中的位置相比处于多个位置时，练习才要求我获得两个列表的乘积。因此，如果此((l1->d + l2->d) % pos == 0))为true，则建立一个新列表。我做了尝试，但我不明白为什么这是错误的。我想念什么？

Nodo *prodotto_pos(Nodo *l1, Nodo *l2, int pos)
{
    Nodo *p;

    if ((l1 == NULL) && (l2 == NULL)) return NULL;
    else if ((l1 != NULL) && (l2 != NULL))
    {
        if (((l1->d + l2->d) % pos) == 0)
        {
            p = newnode();
            p->d = ((l1->d) * (l2->d));
            p->next = prodotto_pos(l1->next, l2->next, pos+1);
        }
    }
    return p;
}

示例：

L1：3-> 4-> 2-> 7-> 5-> 6-> 11-> 16-> 7-> 2-> NULL

L2：0-> 2-> 2-> 6-> 2-> 12-> 2-> NULL

输出：0-> 8-> 72-> NULL

Answer 1

问题是，如果EXEC master.dbo.sp_MSset_oledb_prop N'Microsoft.ACE.OLEDB.12.0', N'AllowInProcess', 1 EXEC master.dbo.sp_MSset_oledb_prop N'Microsoft.ACE.OLEDB.12.0', N'DynamicParameters', 1 exec sp_configure 'Advanced', 1 RECONFIGURE GO exec sp_configure 'Ad Hoc Distributed Queries', 1 RECONFIGURE GO exec sp_configure 'xp_cmdshell', 1 RECONFIGURE GO sp_configure 'show advanced options', 1 RECONFIGURE GO sp_configure 'Ad Hoc Distributed Queries', 1 GO EXEC sp_addlinkedserver @server = N'DATOS', @provider = N'Microsoft.ACE.OLEDB.12.0', @srvproduct = N'OLE DB Provider for ACE', @datasrc = N'C:\SERVIDOR\DATOS\BBDDs.accdb'; GO SELECT id From OpenRowset('Microsoft.ACE.OLEDB.12.0',';Database=C:\SERVIDOR\DATOS\BBDDs.accdb;','SELECT * from Productos') as B GO SELECT * FROM OPENDATASOURCE('Microsoft.ACE.OLEDB.12.0','Data Source=C:\SERVIDOR\DATOS\BBDDs.accdb')...Productos GO错误并且您从函数返回，则不会递归调用该函数突然。

检查下面的product块。

else

Answer 2

您在这里。

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

typedef struct Nodo
{
    int d;
    struct Nodo *next;
} Nodo;

int push_front( Nodo **head, int value )
{
    Nodo *tmp = malloc( sizeof( Nodo ) );
    int success = tmp != NULL;

    if ( success )
    {
        tmp->d = value;
        tmp->next = *head;
        *head = tmp;
    }

    return success;
}

Nodo * prodotto_pos( const Nodo *first, const Nodo *second )
{
    static int pos = 0;

    Nodo *current = NULL;

    if ( first != NULL && second != NULL )
    {
        ++pos;

        if ( ( first->d + second->d ) % pos == 0 )
        {
            current = malloc( sizeof( Nodo ) );
            current->d = first->d * second->d;
            current->next = prodotto_pos( first->next, second->next );
        }
        else
        {
            current = prodotto_pos( first->next, second->next );
        }

        --pos;
    }

    return current;
}

void display( const Nodo *head )
{
    for ( ; head != NULL; head = head->next )
    {
        printf( "%d -> ", head->d );
    }

    puts( "NULL" );
}

int main( void )
{
    Nodo *first = NULL;
    Nodo *second = NULL;

    int a1[] = { 3, 4, 2, 7, 5, 6, 11, 16, 7, 2 };
    const size_t N1 = sizeof( a1 ) / sizeof( *a1 );

    int a2[] = { 0, 2, 2, 6, 2, 12, 2 };
    const size_t N2 = sizeof( a2 ) / sizeof( *a2 );

    for ( size_t i = N1; i != 0; --i )
    {
        push_front( &first, a1[i-1] );
    }

    for ( size_t i = N2; i != 0; --i )
    {
        push_front( &second, a2[i-1] );
    }

    display( first );
    display( second );

    Nodo *product = prodotto_pos( first, second );

    display ( product );
}

程序输出为

3 -> 4 -> 2 -> 7 -> 5 -> 6 -> 11 -> 16 -> 7 -> 2 -> NULL
0 -> 2 -> 2 -> 6 -> 2 -> 12 -> 2 -> NULL
0 -> 8 -> 72 -> NULL

递归计算两个列表之间的乘积（有条件）

2 个答案: