鏈接列表刪除節點後根

我在鏈接列表中練習，我遇到了一個問題。鏈表是尾巴，所以我想刪除位於根節點上方的第一個元素。我不要在任何情況下想刪除根節點。該代碼是：鏈接列表刪除節點後根

struct goods{ 
    char* section; 
    char* pname; 
    int value; 
    int customer_num; 
    int cash; 
    struct goods* next; 
}; 
void delete_goods(int customer_num,struct goods* root) 
{ 
struct goods *current=root; 
struct goods *previous=NULL; 
while(current!=NULL){ 
    if(current->customer_num == customer_num){ 
    if (previous==NULL){ 
     current=current->next; 
     free(root); 
    } 
    else { 
     previous->next=current->next; 
     free(current); 
     current=previous->next; 
    } 
    } 
    else{ 
    previous=current; 
    current=current->next; 
    } 
} 
} 
int main(){ 
    root = malloc(sizeof(struct goods)); 
    root ->next=NULL; 
    printf("give the number of starting customers\n"); 
    scanf("%d",&customers); 
    inform_list(customers);// adds element to list 



     else if(r=='r'&&customers!=0){ 
      printf("removing...\n"); 
      delete_goods(customers,root); 
      customers-=1; 
      printf("customers:\t%d\n",customers); 
      print(); 
     } 
     }

我沒有張貼整個代碼（它包括一些功能添加元素，爲您提供方便，如果你喜歡，我可以做鏈接列表，我需要的方式來解決我的刪除功能所以我滿足上述需求下面列表中的一個示例輸出：
customers: 2 customer: 2 item value: 32 product name: asdasd customer: 1 item value: 43 product name: sdsad customer: 0 item value: 0 product name: (null) 我需要的是我的刪除功能刪除客戶1，如果問那麼顧客2等

來源

2016-04-28 John Vn

如果你不想刪除根節點「在任何情況下「可能在'current = root-> next'和'previous = root'開始遍歷值得考慮。只是說。假設'root'首先被正確地測試了NULL（永遠不會說永遠不會）。並且，考慮到你的限制，我想知道爲什麼'free（root）;'存在於*代碼中的任何位置*。 – WhozCraig

感謝Whoz。該陳述的原因如果（前一個== NULL）{current} = current-> next; free（root）; }'存在我的代碼是，我不能刪除沒有使我的刪除功能崩潰 –

是的...我擔心你做一個明顯錯誤的事情，以防止發生另一個錯誤的事情的狡猾計劃是Baldrick值得:( –

正如其他人所說，你要保留根節點，所以你想從root->next開始（即root將始終爲非空）。

這應該工作[請原諒無償風格清理]：

void 
delete_goods(int customer_num, struct goods *root) 
{ 
    struct goods *current = root->next; 
    struct goods *previous = NULL; 
    struct goods *next; 

    for (; current != NULL; current = next) { 
     next = current->next; 

     if (current->customer_num == customer_num) { 
      if (previous != NULL) 
       previous->next = next; 
      else 
       root->next = next; 

      free(current); 
     } 
     else 
      previous = current; 
    } 
}

這裏有一個稍微更緊湊的版本：

void 
delete_goods(int customer_num, struct goods *root) 
{ 
    struct goods *current = root->next; 
    struct goods *previous = root; 
    struct goods *next; 

    for (; current != NULL; current = next) { 
     next = current->next; 

     if (current->customer_num == customer_num) { 
      previous->next = next; 
      free(current); 
     } 
     else 
      previous = current; 
    } 
}

來源

2016-04-28 20:40:08

感謝您的幫助Craig –

不客氣，我希望引入'next'變量來簡化邏輯的技術變得清晰[以及爲什麼需要]。 –

鏈接列表刪除節點後根

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