从基于位于的索引的C中的链接列表中删除

Question

到目前为止，我已经成功地从第I个索引中删除了所有内容，但我需要能够将第I个之前的节点连接到第I个之后的节点，这就是我目前所拥有的。

LinklistNode* remove_node(LinklistNode* list_head, int index){
if(list_head != NULL){
    LinklistNode* temp;
    if(list_head-> next ==NULL){
        temp = list_head;
        list_head = NULL;
    }
    else{
        LinklistNode* list_pointer = list_head;
        LinklistNode* next_list_pointer = list_pointer->next;
        while(next_list_pointer->next != NULL && index > 0){
            index--;
            list_pointer = next_list_pointer;
            next_list_pointer = next_list_pointer->next;
        }
        temp = next_list_pointer;
        list_pointer->next = NULL;
    }
    free(temp);
   }
   return list_head;
}

Answer 1

要链接索引前后的节点，请将list_pointer->next指向next_list_pointer->next

while(next_list_pointer->next != NULL){
        index--;
        if (index == 0) break;
        list_pointer = next_list_pointer;
        next_list_pointer = next_list_pointer->next;
}
temp = next_list_pointer;
list_pointer->next = temp->next;

Answer 2

为了在第i个元素之前连接节点，请使用list_pointer。有类似的东西…

while(next_list_pointer != NULL || index != 0){
    list_pointer = list_pointer->next;
    next_list_pointer = list_pointer->next;
}
if(next_list_pointer == NULL){
   printf("Invalid index given. Nothing deleted.n");
   //Do not delete otherwise an error will be given
   return list_head;
} else {
   //Valid index was given 
   list_pointer->next = next_list_pointer->next;
   free(next_list_pointer);
   return list_head;
}

这应该几乎涵盖了删除部分，只是一定要包括你的基本情况。

Answer 3

使用指向节点的指针对指针的替代方法：

LinklistNode* remove_node(LinklistNode* list_head, int index)
{
LinklistNode **ppNode = &list_head;
    if(list_head == NULL)
        return NULL;
    while(index--){
        ppNode = &((*ppNode)->next);
        if(*ppNode == NULL)             /* if index not in list */
            return list_head;           /*   just return */
    }
    *ppNode = (*ppNode)->next;          /* delete node */
    return list_head;
}