comparing the elements in a list

If the lists are unsorted then yes two for loops are needed.

If the lists are sorted though... you can do it in one combined loop.

The difference would be that for two unsorted lists each with 10 elements there is a possibility of 100 comparisons to do. With two sorted lists each again with 10 elements the worst case scenario is 20 comparisons.

cotton.ma at 2007-7-12 18:19:20 >

ok i am able to compare the elements of two list. here is the question

Q) write a method to find out if there is an element in the 2nd list that is an element of the first list.

i created a method public boolean retainAll(List xyz). so now when i call this method it gives me an answer true if there is an element and false if there is no duplicates. Does the question requires me to print the duplicates also ? or just to compare whether duplicates exist or not.

Thanks

lrngjavaa at 2007-7-12 18:19:20 >

> No! Don't guess what you have to do. If you are not

> sure, go and ask your teacher.

Actually since the interpretation of vague emotional statements offered in lieu of actual requirements are what I for one spend most of my time doing it might be good to get into the practice now.

Tip: When cornered for failing to meet said "requirements" lean back, sigh heavily and start talking about limitations in any of the following: TCP, clustered indexes, SMTP, multiple inheiritance. The person who gave you the requirements will be so glad to escape from the mind-numbing vortex that you create that they will quickly forgive and forget.

cotton.ma at 2007-7-12 18:19:20 >

////////SPOILER ALERT--\\\\\\\\\\\\\\\\\\\\

THIS IS HOW I WOULD WRITE THE PROGRAM

You clould also add the lists to a HashSet and use

Iterator it = listb.iterator();

Set comp = new HashSet(lista);

ArrayList ar = new list();

while(it.hasNext())

{

Object a = (Object)it.next

if(!comp.add(a))

{

ar.add(a);

}

}

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

{

ar.get(i)

}

>

xpyro_666xa at 2007-7-21 22:06:09 >

Mr Pest. Thanks for the URL. i know about intersection and union. may be my question is too vague. i know set does not allow duplicates.

import java.awt.Color;

import java.util.HashMap;

import java.util.Iterator;

import java.util.Map;

import java.util.Set;

import java.util.*;

public class MapTester

{

public static void main(String[] args)

{

Map<String, Color> favoriteColors

= new HashMap<String, Color>();

favoriteColors.put("Juliet", Color.PINK);

favoriteColors.put("Romeo", Color.GREEN);

favoriteColors.put("Adam", Color.BLUE);

favoriteColors.put("Eve", Color.PINK);

favoriteColors.put("Eve", Color.PINK);

Set<String> keySet = favoriteColors.keySet();

for (String key : keySet)

{

Color value = favoriteColors.get(key);

System.out.println(key + "->" + value);

}

Set set = new TreeSet(keySet);

System.out.println(set);

}

}

But my question is that is there a way to compare 2 list. someone posted a url for retainAll method but its returning true all the time.

like for eg

String id[] = {"Adam", "eve", "Juliet"};

String id1[] = {"Adam", "sony", "pana"};

List a = new ArrayList();

List b = new ArrayList();

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

a.add(id[i]);

for(int i=0; i><id1.length; i++)

b.add(id1[i]);

Now i would like to compare my 2 list to see if there are duplicates or not, rather than converting my list to set is there any other way?

so i tried this

a.retainAll(b);

this returns true whether there is a duplicate or not.

Message was edited by:

lrngjava>

lrngjavaa at 2007-7-21 22:06:09 >

> Mr Pest. Thanks for the URL. i know about

> intersection and union. may be my question is too

> vague. i know set does not allow duplicates.

> But my question is that is there a way to compare 2

> list.

HAHAHAHAHAHAHAHAHAHAHAHAHAHA.

You may want to investigate what "intersection" means.

> someone posted a url for retainAll method

Hell, if you even bothered to read that link you'd see that it says

the intersection method is called retainAll()

TuringPesta at 2007-7-21 22:06:09 >

HAHAHHAHAAHAHAHAHA I am laughing my butt out also. Very funny.

according to the url kevjava posted

retainAll

boolean retainAll(Collection<?> c)Retains only the elements in this list that are contained in the specified collection (optional operation). In other words, removes from this list all the elements that are not contained in the specified collection.

Specified by:

retainAll in interface Collection<E>

Parameters:

c - collection that defines which elements this set will retain.

Returns:

true if this list changed as a result of the call.

Throws:

UnsupportedOperationException - if the retainAll method is not supported by this list.

ClassCastException - if the types of one or more elements in this list are incompatible with the specified collection (optional).

NullPointerException - if this list contains one or more null elements and the specified collection does not support null elements (optional).

NullPointerException - if the specified collection is null.

See Also:

remove(Object), contains(Object)

Am i missing something about intersection or am i blind to see if the intersection is printed?

lrngjavaa at 2007-7-21 22:06:09 >

for each item in listA {

if listB contains current item {

print both lists contain current item

} else {

current item is in listA but not listB

}

}

This requires you to have provided an adequate equals method if your list contain objects of a class you wrote.

Message was edited by:

flounder

floundera at 2007-7-21 22:06:09 >

public boolean hasDuplicateButSideEffect(List a, List b) {

a.retainAll(b);

return a.size() > 0;

}

yeah... simply iterate it and you can add a break when you found one duplicate.

j_shadinataa at 2007-7-21 22:06:14 >