Java 对列表排序（降序）

我怎样才能按降序排列这个列表？它添加数字，但按升序排序，但我希望它是降序的，而不使用任何外部库或任何东西。我想弄明白，但什么也没想出来

private void addNewElement()
{
//getting the input
System.out.print("Please type the number to be added to the list: ");

Integer newValue = null;
while(newValue == null)
{
    try
    {
        newValue = Integer.parseInt(userInput.nextLine());
    }
    catch (final Exception e)
    {
        System.out.println("Wrong value. Please insert new value.");
    }
}

//creating new element based on the input
MyListElement newElement = new MyListElement(newValue);

//if not first
if (firstElement != null)
{
    placeElementInList(newElement);
}
else
{
    firstElement = newElement; //if first
}
}

//if not first
private void placeElementInList(final MyListElement newElement)
{
//if smaller than first
if (newElement.value < firstElement.value)
{
    newElement.nextElement = firstElement;  //new points to first
    firstElement = newElement;              //and becomes first
}
else
{
    MyListElement previousElement = firstElement; //have to remember previous element
    MyListElement elementInList = firstElement.nextElement;  //currently checked.
    while (elementInList != null)
    {
        if (newElement.value < elementInList.value)  //if new element is smaller that currently checked
        {
            break;  //break - put it in current position.
        }
        previousElement = elementInList; //if not, move forward, substitute variables
        elementInList = elementInList.nextElement;
    }
    previousElement.nextElement = newElement;  //set the new element at the proper position
    newElement.nextElement = elementInList;    //
}

---

}

要进行降序排序，只需将所有比较运算符从

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使用Collections.sort和比较器。降序与升序相同，只是翻转了符号。您应该能够通过更改一行代码来更改另一行代码。ifnewElement.value>elementInList.value…好的，但是如果在列表已经完成时我应该按降序排序呢？我是说，在我要打印的时候。如下：MyListElement元素列表=第一个元素；而元素列表！=null{System.out.printelementInList.value+，；elementInList=elementInList.nextElement；}您可以通过交换第一个和最后一个元素、第二个和第二个到最后一个元素等来反转列表，直到到达列表的中间。那么，我应该使用while循环吗？对不起，我真的很困惑

if (newElement.value > firstElement.value)