如何在新的單向鏈表中返回單向鏈表的奇數索引節點？假設第一個節點的索引為 1

Question

當我運行此代碼時，我沒有從編譯器收到錯誤消息，但我無法返回新列表。 我在 MAIN 部分寫錯了代碼嗎？

輸入

10->20->30->40->50->60->70->80->90->100

輸出必須是

10->30->50->70->90

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

typedef struct SinglyLinkedListItem
{
    int data;
    struct SinglyLinkedListItem*next;
}SLLI;


SLLI*OddNodes(SLLI*pHead)
{
    int counter =1;
    SLLI*pTemp=pHead;
    SLLI*pList=NULL;
    while(pTemp != NULL)
    {
        if(counter % 2 != 0)
        {
           if(pList==NULL)
           {
               pList=malloc(sizeof(SLLI));
               pList->data=pTemp->data;
               pList->next=NULL;
           }
           else
           {
               SLLI*pIter=pList;
               SLLI*pNew=malloc(sizeof(SLLI));
               pNew->data=pTemp->data;
               pNew->next=NULL;
               pIter->next=pNew;
               pIter=pIter->next;

           }
        }
        pTemp=pTemp->next;
        counter ++;
    }
    return pList;
}

Answer 1

您總是在更改同一個對象pList->next 。

       else
       {
           pList->next=pTemp;
       }

此外，原始列表不會更改。 因此該函數具有未定義的行為。

對於初學者，您應該通過引用傳遞原始節點的頭部。 否則該函數將處理 head 的副本，副本的任何更改都不會影響原始列表。

這是一個演示程序，展示了如何實現該功能。

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

typedef struct SinglyLinkedListItem
{
    int data;
    struct SinglyLinkedListItem *next;
} SLLI;

SLLI * OddNodes( SLLI **pHead )
{
    int odd = 0;
    SLLI *pList = NULL;
    SLLI **pCurrent = &pList;

    while ( *pHead != NULL )
    {
        if ( odd ^= 1 )
        {
            *pCurrent = *pHead;
            *pHead = ( *pHead )->next;
            ( *pCurrent )->next = NULL;
            pCurrent = &( *pCurrent )->next;
        }
        else
        {
            pHead = &( *pHead )->next;
        }
    }

    return pList;
 }

 int insert( SLLI **pHead, int data )
 {
    SLLI *pCurrent = malloc( sizeof( SLLI ) );
    int success = pCurrent != NULL;

    if ( success )
    {
        pCurrent->data = data;
        pCurrent->next = *pHead;
        *pHead = pCurrent;
    }

    return success;
 }

 void out( SLLI *pHead )
 {
    for ( ; pHead != NULL; pHead = pHead->next )
    {
        printf( "%d -> ", pHead->data );
    }

    puts( "null" );
 }

int main(void) 
{
    const int N = 10;

    SLLI *pHead = NULL;

    for ( int i = N; i != 0; --i )
    {
        insert( &pHead, 10 * i );
    }

    out( pHead );

    SLLI *pSecondHead = OddNodes( &pHead );

    out( pHead );
    out( pSecondHead );

    return 0;
}

函數輸出是

10 -> 20 -> 30 -> 40 -> 50 -> 60 -> 70 -> 80 -> 90 -> 100 -> null
20 -> 40 -> 60 -> 80 -> 100 -> null
10 -> 30 -> 50 -> 70 -> 90 -> null

如果您不打算更改原始列表，那么該函數看起來會更簡單，因為在這種情況下，無需通過引用將指針 pHead 傳遞給該函數。

這是一個演示程序。

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

typedef struct SinglyLinkedListItem
{
    int data;
    struct SinglyLinkedListItem *next;
} SLLI;

SLLI * OddNodes( SLLI *pHead )
{
    int odd = 0;
    SLLI *pList = NULL;
    SLLI **pCurrent = &pList;

    for ( ; pHead != NULL; pHead = pHead->next )
    {
        if ( odd ^= 1 )
        {
            *pCurrent = malloc( sizeof( SLLI ) );
            ( *pCurrent )->data = pHead->data;
            ( *pCurrent )->next = NULL;
            pCurrent = &( *pCurrent )->next;
        }
    }

    return pList;
 }

 int insert( SLLI **pHead, int data )
 {
    SLLI *pCurrent = malloc( sizeof( SLLI ) );
    int success = pCurrent != NULL;

    if ( success )
    {
        pCurrent->data = data;
        pCurrent->next = *pHead;
        *pHead = pCurrent;
    }

    return success;
 }

 void out( SLLI *pHead )
 {
    for ( ; pHead != NULL; pHead = pHead->next )
    {
        printf( "%d -> ", pHead->data );
    }

    puts( "null" );
 }

int main(void) 
{
    const int N = 10;

    SLLI *pHead = NULL;

    for ( int i = N; i != 0; --i )
    {
        insert( &pHead, 10 * i );
    }

    out( pHead );

    SLLI *pSecondHead = OddNodes( pHead );

    out( pHead );
    out( pSecondHead );

    return 0;
}

它的輸出是

10 -> 20 -> 30 -> 40 -> 50 -> 60 -> 70 -> 80 -> 90 -> 100 -> null
10 -> 20 -> 30 -> 40 -> 50 -> 60 -> 70 -> 80 -> 90 -> 100 -> null
10 -> 30 -> 50 -> 70 -> 90 -> null

如果您不了解通過引用使用指針的工作，那么該函數可以如下所示

SLLI * OddNodes( SLLI *pHead )
{
    int odd = 0;
    SLLI *pList = NULL;


    for ( SLLI *pCurrent = pList; pHead != NULL; pHead = pHead->next )
    {
        if ( odd ^= 1 )
        {
            if ( pCurrent == NULL )
            {
                pList = malloc( sizeof( SLLI ) );
                pList->data = pHead->data;
                pList->next = NULL;
                pCurrent = pList;
            }
            else
            {
                pCurrent->next = malloc( sizeof( SLLI ) );
                pCurrent->next->data = pHead->data;
                pCurrent->next->next = NULL;
                pCurrent = pCurrent->next;
            }
        }
    }

    return pList;
}