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Message:
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++)
{
names.add(DATA[i][0]);
times.add(DATA[i][1]);
} //clone the array, and sort
LinkedList sortedTimes = (LinkedList) times.clone();
Collections.sort(sortedTimes); //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
//important.
Object BLANK = new Object();
LinkedList unsortedTimes = (LinkedList) times.clone();
for (Iterator i = sortedTimes.listIterator(); i.hasNext(); )
{
Long time = (Long) i.next();
int index = 0;
while (unsortedTimes.get(index) != time || unsortedTimes.get(index) == BLANK)
index++;
unsortedTimes.set(index, BLANK);
sortedNames.add(names.get(index));
} //showing our results:
for (int i=0; i < sortedNames.size(); i++)
{
System.out.println(sortedNames.get(i) + ":" + sortedTimes.get(i));
}
}
}
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