Question

您好我试图删除链接列表的一部分链接，但我不知道如何删除链接。当我运行它时，链接仍然存在。如果重要，我使用junit来测试函数。

这是我到目前为止所拥有的。

public void removeAt(int k)
{
   Node w = first;
   int counter = 0; 
   if (k<0 ||  k >= size())
   {
       throw new  IndexOutOfBoundsException("Error ");
   }
   else 
   {
       while (w!= null)
       {
         counter++; 
         if (counter == k)
         {
            Node now = w.next; 
            w= now.next; 
         } 
         w=w.next;
       }
   }
   assert check();
}

感谢您的帮助

Answer 1

您需要更改节点的.next字段才能删除节点，例如w.next = w.next.next从列表中删除w.next节点（因为没有任何东西指向它）;一定要检查空指针（如果w.next为null，则w.next.next将抛出异常）。另外，在if块的末尾添加break语句，因为不需要遍历列表的其余部分。

Answer 2

if (counter == k){
    Node now = w.next; 
    w.next= now.next; 
    break;
}

测试一下。

Answer 3

您正在更新本地变量。您需要做的是更新当前节点之前的链接：

 if (k == 0)
    {
      first = first.next ;
      // you also have to free the memory for first.
    }
    else 
    {  
     Node Last = first ;
     w = first.next ;
     counter = 1 ;
     while (w!= null)
     {
       counter++; // Not sure your conventions, but I think this should be at the end
        if (counter == k)
        {
          last.next = w.next ; /// happily skipping w :)
          // remember you have to free w
          break ; // no point in continuing to the end.

        } 
       w=w.next;
     }
    }
    }

Answer 4

您始终需要跟踪上一个节点。而且如果要删除的节点是第一个节点呢？我想你需要改变while块看起来像：

Node l = first;
while (w!= null)
{
    if (counter == k)
    {
        if (w == first)
            first = w.next;
        else    
            l.next = w.next;            
        w.next = null;                  
        break;  
    } 
    l=w;
    w=w.next;
    counter++; 
}

Answer 5

检查以下代码，从链接列表中删除元素

public void delete(T element){

        if(head != null){  // first check your header node is null or not

            // create two references of your linked list 
            Node<T> tmp = head;    // it will hold current value 
            Node<T> tmp1 = head.getNextRef(); // it will hold next value


            while(true){ // iterate through whole linked list

                if(head.getValue() == element){  // if you found element at first place 
                    head = head.getNextRef();   // then point head to next node of it
                    break;
                }

                if(tmp1.getValue()==element){  // to remove node refer to next of tmp1
                    tmp.setNextRef(tmp1.getNextRef());
                    break;   
                }else{
                    tmp = tmp1;  
                    tmp1 = tmp1.getNextRef();

                    if(tmp1==null)
                        break;
                }
            }
        }
    }

删除索引链表

5 个答案: