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November 2001



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Re: Another LinkedList problem...

Posted by Brian Sanders on November 08, 2001 at 12:48 PM

Sorry about the colors on the code listing. Here's a better listing of the same code:

import java.util.*;

public class List
//using actual Long values (rather than Strings) for simplicity
public static Object[][] DATA = {
{ "FirstItem", new Long(50) },
{ "SecondItem", new Long(10) },
{ "ThirdItem", new Long(80) },
{ "FourthItem", new Long(0) },
{ "FifthItem", new Long(10) }

* Sorts the two lists. Preserves the relative order of two events happening at the same time.
public static void main(String[] args)
//initializing the two linked list off of the data in the constant array
LinkedList names = new LinkedList();
LinkedList times = new LinkedList();

for (int i=0; i < DATA.length; i++)

//clone the array, and sort
LinkedList sortedTimes = (LinkedList) times.clone();

//for each time in the sorted times list, find it's index in the unsorted times list.
//this can be used to find the corresponding element in the names list.
LinkedList sortedNames = new LinkedList();

//As we find an index for each time, replace it with a token object to indicate that it's already been used;
//this will allow for duplicate times. We clone the list b/c we assume preserving the original list is
Object BLANK = new Object();
LinkedList unsortedTimes = (LinkedList) times.clone();
for (Iterator i = sortedTimes.listIterator(); i.hasNext(); )
Long time = (Long);
int index = 0;
while (unsortedTimes.get(index) != time || unsortedTimes.get(index) == BLANK)
unsortedTimes.set(index, BLANK);

//showing our results:
for (int i=0; i < sortedNames.size(); i++)
System.out.println(sortedNames.get(i) + ":" + sortedTimes.get(i));


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