从C上的简单链表中删除多个节点答案

【问题标题】：Deleting multiple nodes from simple linked list on C从C上的简单链表中删除多个节点
【发布时间】：2023-03-15 13:26:01
【问题描述】：

我想删除所有与 key 具有相同 idteam 的节点，但它会崩溃...我知道它也应该 free() 内存，但无论如何我认为这应该可以工作：S

//defining the struct
struct players {
   int idplayer;
   int idteam;
   struct players *next;
 };

 struct players *first, *last;

//the function to delete the nodes
void delete(int key){
 struct players *post;
 struct players *pre;
 struct players *auxi;

 auxi = first; //initialization of auxi
 while(auxi != NULL) //this should run the loop till the end of the list?
 {
    if(auxi->idteam == key){ //condition to delete
        if(auxi == first) first = auxi->next; //erase for the case the node is the first one
        else pre->next = post; //erase in the case the node is anywhere else
      }
      pre = auxi; //saves the current value of auxi
      auxi = auxi->next; //increments the position of auxi
      post = auxi->next; //saves the position of the next element
     }
}

【问题讨论】：

stackoverflow.com/a/30578168/971127

标签： c list linked-list nodes erase

【解决方案1】：

  auxi = auxi->next; //increments the position of auxi
  post = auxi->next; //saves the position of the next element

当auxi 变为NULL 时，您最终将执行post = (NULL)->next;，这是访问冲突（崩溃）。

你真的不需要post，只需这样做：

if(auxi->idteam == key){ 
    if(auxi == first) first = auxi->next; 
    else pre->next = auxi->next; // We know auxi is not NULL, so this is safe.
  }

【讨论】：

【解决方案2】：

函数错误。

在这段代码中sn-p

  pre = auxi; //saves the current value of auxi
  auxi = auxi->next; //increments the position of auxi
  post = auxi->next; //saves the position of the next element

后声明

  auxi = auxi->next; //increments the position of auxi

auxi可以等于NULL所以下一条语句

  post = auxi->next; //saves the position of the next element

导致未定义的行为。

但这不是唯一的错误。您还必须正确设置节点last。

你必须释放已删除的节点。

函数可以如下所示

void delete( int key )
{
    struct players *prev = NULL;
    struct players *auxi = first;;

    while ( auxi != NULL )
    {
        if ( auxi->idteam == key )
        {
            struct players *tmp = auxi;

            if ( auxi == first ) first = auxi->next;
            else prev->next = auxi->next;

            if ( auxi == last ) last = prev; 

            auxi = auxi->next; 

            free( tmp );
      }
}

【讨论】：