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了解链表(Java)

[英]Understanding linked lists (Java)

can someone please tell me if I am correct? 有人可以告诉我我是否正确吗? I am studying for a midterm. 我正在学期中。

x is a variable pointing to a linked-list node and not the last node on the list. x是指向链表节点的变量,而不是链表中的最后一个节点。 t points to a new node that is not in the list. t指向不在列表中的新节点。

x.next = t; 
t.next = x.next;

I believe when it comes time to update t.next, x.next is no longer the original node following x, but is instead t itself. 我相信当需要更新t.next时,x.next不再是x之后的原始节点,而是t本身。 So it create a cycle in the list 因此它在列表中创建一个循环

t = x.next 
x = t; 

I believe this does nothing to the list. 我相信这对列表没有任何帮助。

Thank you in advance!! 先感谢您!!

You can also do it threadsafe like this: 您也可以这样进行线程安全:

t.next = x.next; // let t and x point to the SAME next.
x.next = t;  // change the x.next to t(who has the old next)

In this case store node in temp variable. 在这种情况下,将节点存储在temp变量中。 It won't create the cycle. 它不会创建周期。

Object temp = x.next;
x.next = t; 
t.next = temp;

First you have list like this.. 首先,您有这样的列表。

X--->Y----->Z-->

You want to insert a node t after X 您想在X之后插入节点t

Right now t is 现在t

t---->null

Step 1- Now we have temp pointing to X's next 步骤1-现在,我们temp指向X的下一个

x---->y----->z----->
      ^
      |
temp--

Step 2- Now x's next is pointing to t 步骤2-现在x的下一个指向t

x----->t---->

now main list is like this 现在主列表是这样的

temp---->y---->z---->

Step 3- Now t's next is pointing to temp which is only next pointer 步骤3-现在t的下一个指向温度,temp是仅next指针

temp---->y--->z---->
^
|
----------
          |
x---->t---

So resulting list is 所以结果列表是

x--->t---->y---->z----->

You already have object x . 您已经有对象x This probably the current last element of the linked list. 这可能是链表的当前最后一个元素。 Now, you create a new object T and link it as the element after X 现在,您创建一个新对象T并将其链接为X之后的元素

X    // Lets assume X.next == NULL. So linked list looks like this  X -> Null
X.next = T // Now X.next == T and T.Next == NULL, So linked list looks like this  X -> T -> Null. 
T.next = X.next // Now T.next == T. So linked list is  X -> T <->T

This way, when you reach the end of the linked list, it will always return the last element instead of returning NULL . 这样,当您到达链接列表的末尾时,它将始终返回最后一个元素,而不是返回NULL

If you are writing a simple algorithm for this, first you have to create an element and then point its next variable to it self. 如果要为此编写简单的算法,则首先必须创建一个元素,然后将其next变量指向其自身。 <First_element>.next = <First_element> . <First_element>.next = <First_element> So the logic will work for all the instances. 因此,该逻辑将适用于所有实例。

Here is a simple experiment. 这是一个简单的实验。

class Node{
   Node next = null;
   int id =-1;
}

public class LinkedList{
   public static void main (String args[]){
      Node x = new Node();
      x.id = 0;
      x.next = x;

      // Now add a new element
      Node t = new Node();
      t.id  =1;
      x.next = t;
      t.next = x.next; // Now we have a linked list of 2 elements

      Node mynode = x;//First element of linked list
      for(int i =0; i < 3; i++){
        System.out.println(mynode.id);
        mynode = mynode.next;
      }

   }
}

Output: 输出:

0
1
1

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