LinkedList（通用）中的删除方法有什么问题？

Question

我改了 LinkedList 类，但还是不行

LinearNode 类

 public class LinearNode<T>{


  private LinearNode<T> next;
   private T element;

   public LinearNode()
   {
      next = null;
      element = null;
   }

   public LinearNode (T elem)
   {
      next = null;
      element = elem;
   }

   public LinearNode<T> getNext()
   {
      return next;
   }
   public void setNext (LinearNode<T> node)
   {
      next = node;
   }

   public T getElement()
   {
      return element;
   }

   public void setElement (T elem)
   {
      element = elem;
   }
}

我无法找出我的 java 泛型类中删除方法的问题

public void delete(T 元素){

LinearNode<T> previous = list; 
LinearNode<T> current = list;
boolean found = false;

while (!found && current != null)
{
    if (current.getElement ().equals (element)) {
        found = true;
    }
    else {
        previous = current;
        current = current.getNext();
    }
}
//found loop
if (found)//we fount the element
{

    if(current == this.list){
           previous.setNext (null);
           this.last = previous;
       }

   else
       if(current == this.last){
           this.last.setNext(null);
           this.last.equals(previous.getElement()); 

       }
       else{

           previous.setNext(current.getNext());
           current.setNext (null);
       }

this.count--;
}

}

我还有我的驱动程序类，它将从链接列表中删除元素

这里也是驱动类的部分

public void delete(){

        Teacher aTeacher;
        Scanner scan = new Scanner(System.in);
        String number;
        aTeacher = new Teacher();

        System.out.println("Now you can delete teachers from the programme by their number.");
        System.out.println("Please input number:");
        number = scan.nextLine();

        if (aTeacher.getNumber().equals(number)){
        teachers.delete(aTeacher);
        }
        else {
            System.out.println("There are no any teacher with this number.");
        }
    }

Answer 1

我可以在您的代码中看到一些问题。

这个循环有点奇怪

while (current != null && !current.getElement().equals(element))
{
    previous = current;
    current = current.getNext();
    found = true;
}

您不应该在每次迭代时在循环内设置found = true，因为这样您将始终相信在循环完成后您找到了元素。如果您传入您知道列表中存在的值，那么您不会注意到问题。如果您传入不在列表中的值，那么您稍后可能会在代码中看到 current 设置为 null。

我可能会写这个

while (! found && current != null)
{
    if (current.getElement ().equals (element)) {
        found = true;
    }
    else {
        previous = current;
        current = current.getNext();
    }
}

这块也有点奇怪

       if(current == this.last){
           this.last.setNext(null);
           this.last.equals(previous.getElement()); 
       }

这些陈述似乎都没有任何效果。 last.getNext () 的值应该已经是 null。 this.last.equals(previous.getElement()) 只是测试最后一个节点是否等于倒数第二个节点中的元素；该评估应始终为false，并希望没有副作用。

我可能会写这个

       if(current == this.last){
           previous.setNext (null);
           this.last = previous;
       }

最后，虽然删除本身不是问题，但我仍然会在这里彻底，并确保被删除的节点不会保留对列表的任何引用。

所以这个

       previous.setNext(current.getNext());

可能会变成这样

       previous.setNext(current.getNext());
       current.setNext (null);