假设我们的链接列表有 A、C、B 和 D 节点。当你进入第二个 while 循环时说
current = C;
所以使用这个代码：
temp = current; // i.e. temp = C as current = C
current = current.next(); // say current = B now and temp = C
current.setNext(temp); // here B's next is set to C
// but you forgot A's next is C in the example, now since B
// is taking it's place so A's next must point to B
// B's next must point to C and C's next must point to D.
看来你忘记了这些步骤，
之后，当您将 current 移动到下一个节点时，temp 和 current 将交换。但是 temp 之前的那个，即示例中的 A 必须指向 B，它正在与 C 交换。由于 B 之前指向 D，现在交换后 C 必须指向 D（您错过了这部分）并且 B 必须指向C（这就是你在第三行所做的。）
编辑
已添加完整的工作代码以获取更多信息。
import java.io.*;
class Node
{
public Node previous;
public String value;
public Node next;
}
public class LinkedList
{
private BufferedReader br ;
private String str;
private int totalNodes;
private Node current, previous, temp, head, tail;
public LinkedList()
{
br = new BufferedReader(new InputStreamReader(System.in));
current = previous = temp = head = tail = null;
totalNodes = 0;
}
public static void main(String[] args)
{
LinkedList ll = new LinkedList();
ll.menu();
}
private void menu()
{
boolean flag = true;
int choice = 0;
while(flag)
{
System.out.println("--------------------------------------------------");
System.out.println("---------------------MENU-----------------------");
System.out.println("Press 1 : To ADD Node at the END.");
System.out.println("Press 2 : To ADD Node at the BEGINNING.");
System.out.println("Press 3 : To Add Node in BETWEEN the List.");
System.out.println("Press 4 : To SORT the List");
System.out.println("Press 5 : To DISPLAY the List.");
System.out.println("Press 6 : To EXIT the Program.");
System.out.println("--------------------------------------------------");
System.out.print("Please Enter your choice here : ");
try
{
str = br.readLine();
choice = Integer.parseInt(str);
if (choice == 6)
{
flag = false;
}
accept(choice);
}
catch(NumberFormatException nfe)
{
System.out.println("OUCH!, Number Format Exception, entotalNodesered.");
nfe.printStackTrace();
}
catch(IOException ioe)
{
System.out.println("OUCH!, IOException, entotalNodesered.");
ioe.printStackTrace();
}
}
}
private void accept(int choice)
{
switch(choice)
{
case 1:
addNodeToListAtStart();
break;
case 4:
sortListBubble();
break;
case 5:
displayList();
break;
case 6:
System.out.println("Program is Exiting.");
break;
default:
System.out.println("Invalid Choice.\nPlease Refer Menu for further Assistance.");
}
}
private void addNodeToListAtStart()
{
if (head != null)
{
current = new Node();
System.out.print("Enter value for the New Node : ");
try
{
str = br.readLine();
}
catch(NumberFormatException nfe)
{
System.out.println("OUCH!, Number Format Exception, entotalNodesered.");
nfe.printStackTrace();
}
catch(IOException ioe)
{
System.out.println("OUCH!, IOException, entotalNodesered.");
ioe.printStackTrace();
}
current.previous = tail;
current.value = str;
current.next = null;
tail.next = current;
tail = current;
}
else if (head == null)
{
current = new Node();
System.out.print("Enter value for the New Node : ");
try
{
str = br.readLine();
}
catch(NumberFormatException nfe)
{
System.out.println("OUCH!, Number Format Exception, entotalNodesered.");
nfe.printStackTrace();
}
catch(IOException ioe)
{
System.out.println("OUCH!, IOException, entotalNodesered.");
ioe.printStackTrace();
}
current.previous = null;
current.value = str;
current.next = null;
head = current;
tail = current;
}
totalNodes++;
}
private void displayList()
{
current = head;
System.out.println("----------DISPLAYING THE CONTENTS OF THE LINKED LIST---------");
while (current != null)
{
System.out.println("******************************************");
System.out.println("Node ADDRESS is : " + current);
System.out.println("PREVIOUS Node is at : " + current.previous);
System.out.println("VALUE in the Node is : " + current.value);
System.out.println("NEXT Node is at : " + current.next);
System.out.println("******************************************");
current = current.next;
}
}
private boolean sortListBubble()
{
// For Example Say our List is 5, 3, 1, 2, 4
Node node1 = null, node2 = null; // These will act as reference. for the loop to continue
temp = head; // temp is set to the first node.
if (temp == tail || temp == null)
return false;
current = temp.next; // current has been set to second node.
for (int i = 0; i < totalNodes; i++) // this loop will run till whole list is not sorted.
{
temp = head; // temp will point to the first element of the list.
while (temp != tail) // till temp won't reach the second last, as it reaches the last element loop will stop.
{
if (temp != null)
current = temp.next;
while (current != null) // till current is not null.
{
int result = (temp.value).compareToIgnoreCase(current.value);
if (result > 0) // if elment on right side is higher in value then swap.
{
if (temp != head && current != tail) // if nodes are between the list.
{
current.previous = temp.previous;
(temp.previous).next = current;
temp.next = current.next;
(current.next).previous = temp;
current.next = temp;
temp.previous = current;
}
else if (current == tail) // if nodes to be swapped are second last and last(current)
{
temp.next = current.next;
current.previous = temp.previous;
if (temp.previous != null)
(temp.previous).next = current;
else
head = current;
temp.previous = current;
current.next = temp;
tail = temp;
}
else if (temp == head) // if the first two nodes are being swapped.
{
temp.next = current.next;
(current.next).previous = temp;
current.previous = temp.previous;
temp.previous = current;
current.next = temp;
head = current;
}
current = temp.next; // since swapping took place, current went to the left of temp, that's why
// again to bring it on the right side of temp.
}
else if (result <= 0) // if no swapping is to take place, then this thing
{
temp = current; // temp will move one place forward
current = current.next; // current will move one place forward
}
}
if (temp != null)
temp = temp.next;
else // if temp reaches the tail, so it will be null, hence changing it manually to tail to break the loop.
temp = tail;
}
}
return true;
}
}
希望这可能会有所帮助。
问候