Question

我正在尝试为单链接列表实现排序方法。该方法假设通过列表，比较一对节点，并在需要时将其中一个放在前面。它使用另外两种方法： - Remove（）（从列表中删除特定节点） - InsertFront（）（在列表的前面插入一个新节点。这两种方法都可以自行完成，所有内容都可以编译。

    public Link remove(String lastName)
{
    Link current_ = first_;
    Link prior_ = null;
    Link found_ = null;
    while (current_ != null && current_.lastName.compareTo(lastName) != 0)
    {
        prior_ = current_;
        current_ = current_.next;
    }
    if(current_ != null)
    {
        if(prior_ == null)
        {
            found_ = first_;
            System.out.println(current_.next.lastName);
            first_ = current_.next;
        }
        else if(current_ == last_)
        {
            found_ = last_;
            Link hold_ = first_;
            first_ = prior_;
            first_.next = current_.next;
            first_ = hold_;
        }
        else
        {
            found_ = current_;
            Link hold_ = first_;
            first_ = prior_;
            first_.next = current_.next;
            first_ = hold_;
        }
    }
    return found_;
}

    public void insertFront(String f, String m, String l)
{
    Link name = new Link(f, m, l);
    if (isEmpty())
    {
        System.out.println("Adding first name");
        first_ = name;
        last_ = name;

    }
    else
    {
        System.out.println("Adding another name");
        Link hold_ = first_;
        first_ = last_;
        first_.next = name;
        last_ = first_.next;
        first_ = hold_;
    }
}

我试图修复它，但我总是遇到两个不同的问题：

它可以工作，但不会重新插入链接。

    public void Sort()
{
Link temp_;
boolean swapped_ = true;
while (swapped_ == true)
{
    swapped_ = false;
    Link current_ = first_;
    String comp_ = current_.lastName;

    while (current_ != null && current_.lastName.compareTo(comp_) >= 0)
    {
        current_ = current_.next;
    }

    if (current_ != null)
    {
        temp_ = remove(current_.lastName);
        insertFront(temp_.firstName, temp_.middleName, temp_.lastName);
        swapped_ = true;
    }
  }
}

我收到Null Pointer Exception。

Exception in thread "main" java. lang. NullPointerException
at List $ Link . access $000(List . java : 25)
at List . Sort ( List . java:165)
at main( java :79)

Java结果：1

DeBugging导致：

Listening on javadebug
Not able to submit breakpoint LineBreakpoint LinearGenericSearch.java : 28, reason: The breakpoint is set outside of any class.
Invalid LineBreakpoint LinearGenericSearch.java : 28
User program running
Debugger stopped on uncompilable source code.
User program finished


    public void Sort()
        Link temp_;     
    boolean swapped_ = true;
    while (swapped_ == true)
    {
            Link current_ = first_;
            swapped_ = false;
            while (current_.next != null)
            {

                if((current_.lastName.compareTo(current_.next.lastName)) > 0)
                {
                    temp_ = remove(current_.next.lastName);
                    insertFront(temp_.firstName, temp_.middleName, temp_.lastName);
                    swapped_ = true;
                }
                current_ = current_.next;
            }
        }
    }

我的问题是：我做错了什么？关于下次如何避免这种情况的任何建议？

Answer 1

在java中你不必手动完成所有这些！只需使用LinkedList，即代码

 compareTo()

实施

 Comparable Interface

（http://docs.oracle.com/javase/6/docs/api/java/lang/Comparable.html）并致电

 Collections.sort(List list, Comparator c)

（http://docs.oracle.com/javase/6/docs/api/java/util/Collections.html）按你想要的方式排序。

为什么链接列表的Sort方法不起作用？

1 个答案: