c中的指针:删除链表中每一个元素的函数

Question

我想写一个函数，它获取一个指向链表头部的指针，并每隔一秒从列表中删除一个成员。list是一个类型为element:的链接元素

typedef struct element{
    int num;
    struct element* next;
}element;

我对所有这些指针算术都是新手，所以我不确定我写得是否正确：

 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;
            }
        }
    }

自从我不断发现错误以来，我一直在更改它。你能指出任何可能的错误吗？

Answer 1

如果您将链表视为节点对，则算法会简单得多。循环的每次迭代都应该处理两个节点——head和head->next，并在退出时使head等于head->next->next。同样重要的是，如果要将中间节点从列表中删除，请不要忘记删除它，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);
}

Answer 2

用递归的术语来可视化这个问题可能是最简单的，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，并导致编译器出现堆栈溢出，而编译器没有对其进行优化。相反，dasblinkenlight的答案包含了您应该实际使用的代码。