c中的指针：函数，删除链表中的每一个元素

我想写一个函数，它获取一个指向链表头的指针，并每隔一秒从链表中删除一个成员。该列表是一个链接元素，类型为元素：

我对所有这些指针算法都是新手，所以我不确定我写的是否正确：

 void deletdscnds(element* head) {
    element* curr;
    head=head->next; //Skipping the dummy head//

    while (head!=NULL) {
        if (head->next==NULL) 
            return;

            else {
                curr=head;
                head=head->next->next; //worst case I'll reach NULL and not a next of a null//
                curr->next=head;
            }
        }
    }

---

我不断地修改它，因为我不断地发现错误。您能指出任何可能的错误吗？

如果您从节点对的角度考虑链表，算法会简单得多。循环的每次迭代都应处理两个节点-

head

和

head->next

，并在退出时使

head

等于

head->next->next

。同样重要的是，如果要从列表中删除中间节点，不要忘记删除它，否则会出现内存泄漏

while (head && head->next) {
    // Store a pointer to the item we're about to cut out
    element *tmp = head->next;
    // Skip the item we're cutting out
    head->next = head->next->next;
    // Prepare the head for the next iteration
    head = head->next;
    // Free the item that's no longer in the list
    free(tmp);
}

---

用递归的术语来形象化这个问题可能是最直接的，比如：

// outside code calls this function; the other functions are considered private
void deletdscnds(element* head) {
  delete_odd(head);
}

// for odd-numbered nodes; this won't delete the current node
void delete_odd(element* node) {
  if (node == NULL)
    return; // stop at the end of the list
  // point this node to the node two after, if such a node exists
  node->next = delete_even(node->next);
}

// for even-numbered nodes; this WILL delete the current node
void delete_even(element* node) {
  if (node == NULL)
    return NULL; // stop at the end of the list
  // get the next node before you free the current one, so you avoid
  // accessing memory that has already been freed
  element* next = node->next;
  // free the current node, that it's not needed anymore
  free(node);
  // repeat the process beginning with the next node
  delete_odd(next);
  // since the current node is now deleted, the previous node needs
  // to know what the next node is so it can link up with it
  return next;
}

至少对我来说，这有助于澄清每一步需要做什么

---

我不建议实际使用这种方法，因为在C语言中，递归算法可能会占用大量RAM，并导致编译器无法优化堆栈溢出。相反，dasblinkenlight的答案包含了您应该实际使用的代码。

Yikes！在你这样做一段时间后，你将站在膝盖深的泄漏记忆！您没有删除任何内容…您刚刚丢失了它。我应该在释放它之前使用“free”功能吗？您必须在删除curr之前获取curr->next的值。

// outside code calls this function; the other functions are considered private
void deletdscnds(element* head) {
  delete_odd(head);
}

// for odd-numbered nodes; this won't delete the current node
void delete_odd(element* node) {
  if (node == NULL)
    return; // stop at the end of the list
  // point this node to the node two after, if such a node exists
  node->next = delete_even(node->next);
}

// for even-numbered nodes; this WILL delete the current node
void delete_even(element* node) {
  if (node == NULL)
    return NULL; // stop at the end of the list
  // get the next node before you free the current one, so you avoid
  // accessing memory that has already been freed
  element* next = node->next;
  // free the current node, that it's not needed anymore
  free(node);
  // repeat the process beginning with the next node
  delete_odd(next);
  // since the current node is now deleted, the previous node needs
  // to know what the next node is so it can link up with it
  return next;
}