我必须为我的编程课程创建一个Linked List程序。它工作,每次插入一个数字,它都放在列表的开头。现在,我的老师希望我们采用我们的链接列表程序并按升序对数字进行排序。我完全迷失了如何做到这一点。谁能指出我正确的方向? 这是我的列表代码:
public class SortedList {
private DoubleNode head = null;
private int listLength;
public static void main(String[] args) {
SortedList list = new SortedList();
list.insert(6);
list.insert(7);
System.out.println(list.toString());
}
public void insert(double value) {
head = new DoubleNode(value, head);
listLength++;
}
public String toString() {
String answer = "[ ";
for (DoubleNode current = head; current != null; current = current
.getLink()) {
answer += current.getData() + " ";
}
answer += "]";
return answer;
}
public int find(double value) {
if (listLength == 0)
return -1;
int pos = 1;
for (DoubleNode current = head; current != null; current = current.getLink()) {
if (current.getData() == value)
return pos;
pos++;
}
return -1;
}
public int size() {
return listLength;
}
public boolean removeAt(int index) {
if (index < 1 || index > listLength)
return false;
if (index == 1) {
if (head != null) {
head = head.getLink();
listLength--;
}
return true;
}
DoubleNode current = head;
for (int i = 1; i < (index - 1); i++) {
if (current.getLink() == null)
return false;
current = current.getLink();
}
current.setLink(current.getLink().getLink());
listLength--;
return true;
}
}
这是我老师给出的节点的代码:
// File: DoubleNode.java based on the DoubleNode class by Michael Main
/**************************************************************************
* DoubleNode provides a node for a linked list with double data in each node.
*
* @note
* Lists of nodes can be made of any length, limited only by the amount of
* free memory in the heap.
*
* @author Michael Main
* shortened by Beth Katz and Stephanie Elzer to be only the basics
*
* @version
* February 2007
***************************************************************************/
public class DoubleNode
{
// Invariant of the DoubleNode class:
// 1. The node's double data is in the instance variable data.
// 2. For the final node of a list, the link part is null.
// Otherwise, the link part is a reference to the next node of the list.
private double data;
private DoubleNode link;
/**
* Initialize a node with a specified initial data and link to the next
* node. Note that the initialLink may be the null reference, which
* indicates that the new node has nothing after it.
* @param initialData
* the initial data of this new node
* @param initialLink
* a reference to the node after this new node--this reference may be
* null to indicate that there is no node after this new node.
* @postcondition
* This node contains the specified data and link to the next node.
**/
public DoubleNode(double initialData, DoubleNode initialLink)
{
data = initialData;
link = initialLink;
}
/**
* Accessor method to get the data from this node.
* @param - none
* @return
* the data from this node
**/
public double getData( )
{
return data;
}
/**
* Accessor method to get a reference to the next node after this node.
* @param - none
* @return
* a reference to the node after this node (or the null reference if
* there is nothing after this node)
**/
public DoubleNode getLink( )
{
return link;
}
/**
* Modification method to set the data in this node.
* @param newData
* the new data to place in this node
* @postcondition
* The data of this node has been set to newData.
**/
public void setData(double newData)
{
data = newData;
}
/**
* Modification method to set the link to the next node after this node.
* @param newLink
* a reference to the node that should appear after this node in the
* linked list (or the null reference if there is no node after this node)
* @postcondition
* The link to the node after this node has been set to newLink. Any other
* node (that used to be in this link) is no longer connected to this node.
**/
public void setLink(DoubleNode newLink)
{
link = newLink;
}
}
答案 0 :(得分:3)
想法
1)在最简单的情况下,列表已经排序:
- &GT;甲
2)现在,考虑“下一个”情况(即将1个新元素添加到大小为1的列表中)
- &GT; A [现在,我将尝试添加C]
你可以简单地检查C是否&gt;比A,在这种情况下,你在末尾添加“C”( - &gt; A-&gt; C)
3)我们可以概括案例(2):在任何后续情况下,你必须走下列表,直到你“看到”一个新的节点,即>而不是你要插入的那个。
- &GT; A - &gt; C [加B]
check 1: A (B > A)
check 2: C (B < C) !
这意味着我们可以按如下方式替换链接:
替换A - &gt; C有2个新链接,1个来自A - &gt; B,以及另一个来自B - >下进行。
以这种方式插入gaurantees您的列表仍然排序。
<强>具体强>
因此,您必须修改应用程序的insert(..)方法,从列表的开始处开始,检查每个DoubleNode,向下走,并“记住”即存储前一个DoubleNode,直到它或者到达列表的末尾---或者它看到最后一个节点是&lt;而不是新节点,并且当前节点是>比新节点。