使用java

时间:2016-11-30 10:20:54

标签: java sorting linked-list mergesort

我想使用java,只使用没有任何数组的链表实现合并排序。但我陷入了逻辑错误;我的代码消除了一些输入并对遗骸进行了排序。我已经应用了三个课程:DivideMergMergSort,如下所示:

public class Divide {

    List firstList = new List();
    List secondList = new List();

    public int GetLength(List list) {
        int Length = 0;
        Link temp = new Link();
        temp.next = list.head.next;
        while (temp.next != null) {
            temp.next = temp.next.next;
            Length++;
        }

        return Length;
    }

    public List rightSide(List list) {
        int Length = GetLength(list);
        Link temp = new Link();
        temp.next = list.head.next;

            for (int i = 1; i <= Math.floor(Length / 2); i++) {
                firstList.Insert(temp.next.data);
                temp.next = temp.next.next;
            }
            return firstList;
    }

    public List leftSide(List list) {
        int Length = GetLength(list);
        Link temp = new Link();
        temp.prev = list.head.prev;

            for (int i = 1; i <= Math.ceil(Length / 2); i++) {
                secondList.Insert( temp.prev.data);
                temp.prev =  temp.prev.prev;
            }
            return secondList;
        }
    }

合并:

public class Merg {

    public List MergedList = new List();
    private  List Temp = new List();

    public List Merg (List one, List two)
    {
        Link onelink = new Link();
        Link twolink = new Link();

        onelink.next = one.head.next;
        twolink.next = two.head.next;

        while (onelink.next!= null || twolink.next!= null)
        {

            if(onelink.next!= null && twolink.next != null)
            {
              if(onelink.next.data < twolink.next.data)
              {
                  Temp.Insert(onelink.next.data);
                  onelink.next = onelink.next.next;
              }
              else
              {
                  Temp.Insert(twolink.next.data);
                  twolink.next = twolink.next.next;
              }

            }

            if (onelink.next != null && twolink.next == null)
            {
                Temp.Insert(onelink.next.data);
                onelink.next = onelink.next.next;
            }
            if (twolink.next != null && onelink.next == null)
            {
                Temp.Insert(twolink.next.data);
                twolink.next = twolink.next.next;
            }

        }

        if (Temp.head.next.data > Temp.head.prev.data)
        {
            while (Temp.head.next != null)
            {
                MergedList.Insert(Temp.head.next.data);
                Temp.head.next = Temp.head.next.next;
            }
        }

        else
        {
            MergedList.head.next = Temp.head.next;
        }


        return MergedList;
    }

}

MergSort:

public class MergSort {

    public List mergSort (List list)
    {
        Divide divide = new Divide();
        Merg merg = new Merg();

        if (divide.GetLength(list) > 1) {
            return merg.Merg(mergSort(divide.leftSide(list)), mergSort(divide.rightSide(list)));
        }
        else
        {
            return list;
        }
    }
}

虽然我也写了linklist类,但我认为那里没有问题。 (但如果有必要,我会提到它们) 现在,当我导入一些输入时:{100,3,1,7,6}输出为:3,6,7,100。 (1已被淘汰!)或另一个例子:{100,3,1}输出为:1,100(其中是3 ??)

我想知道是否有人可以帮助我...

链接:

public class Link {
    public double data;
    public Link prev;
    public Link next;
}

列表:

public class List {
    public Link head = new Link();

    public void setHead(Link head) {
        this.head = head;
        head.prev = null;
        head.next = null;
    }

    public void Insert(double x)
    {
        Link link = new Link();
        link.data = x;
        if (head.next == null)
        {
            head.next = link;
            head.prev = link;
        }
        else
        {
            head.next.prev = link;
            link.next = head.next;
            head.next = link;
        }
    }

    public Link delete()
    {
        Link temp = new Link();
        temp = head;
        head = head.next;
        return temp;
    }

    public Link search(double x)
    {
        Link link = new Link();
        link.next = head;
        while (link.next.data != x)
        {
            link.next = link.next.prev;
        }

        return link.next;
    }
}

1 个答案:

答案 0 :(得分:0)

我已经调试了很多,但我无法理解你的Divide课程。你根本不需要它。

public class MergSort {

    public List mergSort(List list) {
        Divide divide = new Divide();
        Merg merg = new Merg();
        int n = list.getLength();
        if (n > 1) {
//            return merg.Merg(mergSort(divide.leftSide(list)), mergSort(divide.rightSide(list)));
            Link cursor = list.head.next;
            List left = new List();
            List right = new List();
            // for (i = 0; ....... i will be 0 if head is not dummy
            for (int i = 1; cursor != null; i++) {
                if (i <= n/2)
                    left.Insert(cursor.data);
                else
                    right.Insert(cursor.data);
                cursor = cursor.next;
            }
            left = mergSort(left);
            right = mergSort(right);

            return merg.Merg(left, right);

        } else {
            return list;
        }
    }
}

getLength()课程中创建List方法。它应该在List类中,因为您需要列表的长度。

public int getLength() {
    int Length = 0;
    Link temp = head.next;
    while (temp != null) {
        temp = temp.next;
        Length++;
    }

    return Length;
}

另外,将您的Link课程重命名为Node。因为Link表示对象之间的链接或节点之间的边缘。你真正想要的是Node。这令人困惑。

List课程中,您已定义head,如下所示:

public Link head = new Link();

我不明白你为什么需要dummy headed linked list

带有虚拟头的链表基本上是没有值的头,因此是虚拟的。

enter image description here

由于你的假人头,你正在Divide i = 1班开始你的for循环。为什么复杂的事情?如果你真的需要一个假头,那么保留它或者像这样定义:

public Link head;
// don't instantiate the head. It's just a reference

此外,您似乎创建了一个半圆链表。我的意思是head.prev指向列表中的最后一个链接/节点。如果您真的想要一个循环链表,那么请将tail.next指向头部。

这是循环链表的样子:

enter image description here

或者,如果您不想要循环列表,请不要将head.prev指向最后一个对象。

Insert()中的List.java方法:

if (head.next == null)
{
    head.next = link;
    // remove head.prev = link;
}