Question

我目前正在使用节点制作一个链接的List程序（不是我知道的任何其他方式）我遇到了一个关于创建一个深拷贝并用我的~List（）删除所有节点和Sentinels的问题。删除节点不是问题，但是哨兵是因为第一个没有分配索引值。

List::~List()
{
  for(size_t i=0; i<size; i++)
    {
      _setCurrentIndex(i);
      if(current && curent->next == NULL)
    {
      Node *temp = current->next;
      delete temp;
      delete current;
    }
      else
    {
      Node *old = current;
      current = current->next;
      delete old;
    }
    }
}

List::List(const List & orig)
{
for(size_t i=0; i<size; i++)
{
 if(i==0)
  {
   Node *copyFront = new Node; //the first sentinel
   copyFront->data = orig.front->data; //front is defined in private in list.h
   copyFront->prev = NULL; // it is defined as a Node (same for rear)
  }
 else if(0<=i && i<size) //put in i<size b/c 0<=i would always be true
  {
   _setCurrentIndex(i) //sets what current is and currentIndex which pts to diff Nodes
   Node *copy = new Node;
   copy->data = current->data; 
   copy->next = current->next;
   current = current->next;
  }
 else if(i+1 == size)
  {
   Node *copyRear = new Node; //making the last sentinel, but it has to be
   copyRear->data = orig.rear->data; //after data Node
   copyRear->next = NULL;
  }
 }
}

我正在寻求关于此代码的建议和评论，如果出现严重错误，可以继续下一步或改变什么！

Answer 1

链接列表是允许任何类型的变量位于其中的模板。我诚实地认为，您最好使用需要std::list头文件的#include <list>。

当然，如果您真的想要自己编写链表类的经验，那么以下代码会对列表进行深层复制：

List::List( const List& other) {
    if( other.head_ != nullptr) {
        head_ = new Node( other.head_->item_);  // copy first node

        assert( head_ != nullptr);  // ensure that the memory was allocated correctly

        // copy the rest of the list
        Node* pnew = head_;
        // loop through the list until you reach the end (i.e. a node that's nullptr)
        for( Node* porig( other.head_->next_); porig != nullptr; porig = porig->next_) {
            // assign the next node in the destination list to the next node in the paramter's list
            pnew->next_ = new Node( porig->item_);
            assert( pnew->next_ != nullptr);  // ensure that the memory was allocated correctly
            pnew = pnew->next_;  // move onto the newly created node in the destination list
        }
    }
    else
        // if the parameter is empty then the destination list will be empty as well
        head_ = nullptr;
}

对于析构函数，您只需要在列表中运行删除节点：

List::~List() {
    while( head_ != nullptr) {  // keep looping until the list gets to the end
        // make a second pointer to the node you are about to delete (so you don't lose track of it)
        Node* pn( head_);
        // move the head_ onto the next node essentially "removing" the first node from your list
        head_ = head_->next_;
        // delete the node that you've just "removed" from your list
        delete pn;
    }
}

我试图澄清任何可能不清楚的内容。

使用节点的深层复制和解构器标记链表

1 个答案: