c - 在这个函数中正确的指针没有返回到调用函数

Question

问题是当我在删除节点后打印列表时，打印函数打印 0 代替删除的节点.....但我希望它不打印任何内容。

// the function call is delete_from_key(&head,i);
//node is struct linked_list
void delete_from_key(node *head, int key)
{
    node *new, *temp;
    if(head == NULL)
    {
        printf("nothing to delete . the list is empty.\n");
        return;
    }
    else if(head->num == key)
    {
        temp=head;
        head=temp->next;
        free(temp);
        return;
    }
    new=search_key(head, key);//search_key returns pointer node holding the key.
    if(new != NULL)
    {
        temp = new;
        new = new->next;
        free(temp);
        return;
    }
}

例子:

列表是1->2->3->4->5->
如果我用键 2 调用这个函数
预期的输出是 1-> 3-> 4-> 5->
相反，实际输出是 1-> 0-> 3-> 4-> 5->

Answer 1

您需要更新前一个节点的 next 以及它的 next 正在被删除。

prev_node=search_key(head, key);//search_key returns pointer to the node just before the node holding the key.

if(prev_node!= NULL)
{
    // prev_node->next is the one that needs to be deleted
    temp = prev_node->next; 

    //Make the prev node point to the next node of the one that's getting deleted
    prev_node->next = temp->next; 

    free(temp);
    return;
}