Question

我必须编写一个接收两个排序列表的递归函数：

typedef struct listNode {
    int* dataPtr;
    struct listNode* next;
} ListNode;

typedef struct list
{
    ListNode* head;
    ListNode* tail;
} List;

并将它们合并到一个排序的列表中。

我写了这些函数：

void mergeRec(ListNode* head1, ListNode* head2, ListNode* mergedList)
{
    if (head1 == NULL && head2 == NULL)
        return;
    else if (head1 == NULL) {
        mergedList->next = head2;
        head2 = head2->next;
    }
    else if (head2 == NULL) {
        mergedList->next = head1;
        head1 = head1->next;
    }
    else if (*(head1->dataPtr) > *(head2->dataPtr)) {
        mergedList->next = head1;
        head1 = head1->next;
    }
    else
    {
        mergedList->next = head2;
        head2 = head2->next;
    }

    mergeRec(head1, head2, mergedList->next);
}

List merge(List lst1, List lst2)
{
    List mergedList;
    makeEmptyList(&mergedList);
    mergeRec(lst1.head, lst2.head, mergedList.head);
    return mergedList;
}

现在，递归函数存在的问题是，在合并列表指向空值的第一个调用中，显然，当我编写类似mergeList->next之类的东西时，会遇到运行错误。

我试图通过在递归中添加以下条件来解决它：

    if (mergedList == NULL)
    {
        if (*(head1->dataPtr) > *(head2->dataPtr))
        {
            mergedList = head1;
            head1 = head1->next;
        }
        else
        {
            mergedList = head2;
            head2 = head2->next;
        }
    }

但我收到此错误：

“在q2d.exe中的0x00661EB9处引发了异常：0xC0000005：访问冲突写入位置0x01000F48。”

我无法说出问题或解决方法。非常感谢您的帮助。

谢谢！

Answer 1

对于初学者来说，目前还不清楚为什么在此结构中存在类型为int *的数据成员

typedef struct listNode {
    int* dataPtr;
    struct listNode* next;
}List;

不仅仅是类型int

typedef struct listNode {
    int data;
    struct listNode* next;
}List;

尽管如此，函数merge和mergeRec无效，因为它们处理list以及指针list1.head，list2.head和mergedList.head的值的副本。

List merge(List lst1, List lst2)
mergeRec(lst1.head, lst2.head, mergedList.head);

此外，指针list1.tail，list2.tail和mergedList.tail被忽略。

我可以建议下面的演示程序中显示以下解决方案。

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

typedef struct listNode 
{
    int *dataPtr;
    struct listNode *next;
} ListNode;

typedef struct list
{
    ListNode *head;
    ListNode *tail;
} List;

void makeEmpty( List *list )
{
    list->head = list->tail = NULL;
}

int push( List *list, int data )
{
    ListNode *current = malloc( sizeof( ListNode ) );
    int success = current != NULL;

    if ( success )
    {
        current->dataPtr = malloc( sizeof( int ) );
        success = current->dataPtr != NULL;

        if ( success )
        {
            *current->dataPtr = data;
            current->next = NULL;

            if ( list->head == NULL )
            {
                list->head = list->tail = current;
            }
            else
            {
                list->tail = list->tail->next = current;
            }
        }
        else
        {
            free( current );
            current = NULL;
        }
    }

    return success;
}

List merge( List *first, List *second )
{
    List result;
    makeEmpty( &result );

    if ( ( second->head != NULL ) && 
         ( first->head == NULL || *second->head->dataPtr < *first->head->dataPtr ) )
    {
        result.head = result.tail = second->head;
        second->head = second->head->next;
        if ( second->head == NULL ) second->tail = NULL;
    }
    else if ( first->head != NULL )
    {
        result.head = result.tail = first->head;
        first->head = first->head->next;
        if ( first->head == NULL ) first->tail = NULL;
    }

    if ( !( first->head == NULL && second->head == NULL ) )
    {
        List tmp = merge( first, second );
        result.head->next = tmp.head;
        result.tail = tmp.tail;
    }

    return result;
}

void output( const List *list )
{
    for ( const ListNode *current = list->head; current != NULL; current = current->next )
    {
        printf( "%d ", *current->dataPtr );
    }

    puts( "NULL" );
}

int main(void) 
{
    List even_numbers;
    List odd_numbers;

    makeEmpty( &even_numbers );
    makeEmpty( &odd_numbers );

    const int N = 10;

    for ( int i = 0; i < N; i++ ) 
    {
        i % 2 == 0 ? push( &even_numbers, i )
                   : push( &odd_numbers, i );
    }

    printf( "even_numbers: " ); output( &even_numbers );
    printf( "odd_numbers:  " ); output( &odd_numbers );

    List all_numbers = merge( &even_numbers, &odd_numbers );

    printf( "all_numbers:  " ); output( &all_numbers );
    printf( "even_numbers: " ); output( &even_numbers );
    printf( "odd_numbers:  " ); output( &odd_numbers );

    return 0;
}

程序输出为

even_numbers: 0 2 4 6 8 NULL
odd_numbers:  1 3 5 7 9 NULL
all_numbers:  0 1 2 3 4 5 6 7 8 9 NULL
even_numbers: NULL
odd_numbers:  NULL

将两个排序的链表合并为一个链表（递归）

1 个答案: