Related
For lists, we use the Collections.sort(List) method. What if we want to sort a HashSet?
A HashSet does not guarantee any order of its elements. If you need this guarantee, consider using a TreeSet to hold your elements.
However if you just need your elements sorted for this one occurrence, then just temporarily create a List and sort that:
Set<?> yourHashSet = new HashSet<>();
...
List<?> sortedList = new ArrayList<>(yourHashSet);
Collections.sort(sortedList);
Add all your objects to the TreeSet, you will get a sorted Set. Below is a raw example.
HashSet myHashSet = new HashSet();
myHashSet.add(1);
myHashSet.add(23);
myHashSet.add(45);
myHashSet.add(12);
TreeSet myTreeSet = new TreeSet();
myTreeSet.addAll(myHashSet);
System.out.println(myTreeSet); // Prints [1, 12, 23, 45]
Update
You can also use TreeSet's constructor that takes a HashSet as a parameter.
HashSet myHashSet = new HashSet();
myHashSet.add(1);
myHashSet.add(23);
myHashSet.add(45);
myHashSet.add(12);
TreeSet myTreeSet = new TreeSet(myHashSet);
System.out.println(myTreeSet); // Prints [1, 12, 23, 45]
Thanks #mounika for the update.
Java 8 way to sort it would be:
fooHashSet.stream()
.sorted(Comparator.comparing(Foo::getSize)) //comparator - how you want to sort it
.collect(Collectors.toList()); //collector - what you want to collect it to
*Foo::getSize it's an example how to sort the HashSet of YourItem's naturally by size.
*Collectors.toList() is going to collect the result of sorting into a List the you will need to capture it with List<Foo> sortedListOfFoo =
You can use a TreeSet instead.
Use java.util.TreeSet as the actual object. When you iterate over this collection, the values come back in a well-defined order.
If you use java.util.HashSet then the order depends on an internal hash function which is almost certainly not lexicographic (based on content).
Just in-case you don't wanna use a TreeSet you could try this using java stream for concise code.
set = set.stream().sorted().collect(Collectors.toCollection(LinkedHashSet::new));
You can use Java 8 collectors and TreeSet
list.stream().collect(Collectors.toCollection(TreeSet::new))
Based on the answer given by #LazerBanana i will put my own example of a Set sorted by the Id of the Object:
Set<Clazz> yourSet = [...];
yourSet.stream().sorted(new Comparator<Clazz>() {
#Override
public int compare(Clazz o1, Clazz o2) {
return o1.getId().compareTo(o2.getId());
}
}).collect(Collectors.toList()); // Returns the sorted List (using toSet() wont work)
Elements in HashSet can't be sorted. Whenever you put elements into HashSet, it can mess up the ordering of the whole set. It is deliberately designed like that for performance. When you don't care about the order, HashSet will be the most efficient set for frequent insertions and queries.
TreeSet is the alternative that you can use. When you iterate on the tree set, you will get sorted elements automatically.
But it will adjust the tree to try to remain sorted every time you insert an element.
Perhaps, what you are trying to do is to sort just once. In that case, TreeSet is not the most efficient option because it needs to determine the placing of newly added elements all the time. Use TreeSet only when you want to sort often.
If you only need to sort once, use ArrayList. Create a new list and add all the elements then sort it once. If you want to retain only unique elements (remove all duplicates), then put the list into a LinkedHashSet, it will retain the order you have already sorted.
List<Integer> list = new ArrayList<>();
list.add(6);
list.add(4);
list.add(4);
list.add(5);
Collections.sort(list);
Set<Integer> unique = new LinkedHashSet<>(list); // 4 5 6
Now, you've gotten a sorted set if you want it in a list form then convert it into list.
You can use TreeSet as mentioned in other answers.
Here's a little more elaboration on how to use it:
TreeSet<String> ts = new TreeSet<String>();
ts.add("b1");
ts.add("b3");
ts.add("b2");
ts.add("a1");
ts.add("a2");
System.out.println(ts);
for (String s: ts)
System.out.println(s);
Output:
[a1, a2, a3, a4, a5]
a1
a2
b1
b2
b3
In my humble opinion , LazerBanana's answer should be the top rated answer & accepted because all the other answers pointing to java.util.TreeSet ( or first convert to list then call Collections.sort(...) on the converted list ) didn't bothered to ask OP as what kind of objects your HashSet has i.e. if those elements have a predefined natural ordering or not & that is not optional question but a mandatory question.
You just can't go in & start putting your HashSet elements into a TreeSet if element type doesn't already implement Comparable interface or if you are not explicitly passing Comparator to TreeSet constructor.
From TreeSet JavaDoc ,
Constructs a new, empty tree set, sorted according to the natural
ordering of its elements. All elements inserted into the set must
implement the Comparable interface. Furthermore, all such elements
must be mutually comparable: e1.compareTo(e2) must not throw a
ClassCastException for any elements e1 and e2 in the set. If the user
attempts to add an element to the set that violates this constraint
(for example, the user attempts to add a string element to a set whose
elements are integers), the add call will throw a ClassCastException.
That is why only all Java8 stream based answers - where you define your comparator on the spot - only make sense because implementing comparable in POJO becomes optional. Programmer defines comparator as and when needed. Trying to collect into TreeSet without asking this fundamental question is also incorrect ( Ninja's answer). Assuming object types to be String or Integer is also incorrect.
Having said that, other concerns like ,
Sorting Performance
Memory Foot Print ( retaining original set and creating new sorted sets each time sorting is done or wish to sort the set in - place etc etc )
should be the other relevant points too. Just pointing to API shouldn't be only intention.
Since Original set already contains only unique elements & that constraint is also maintained by sorted set so original set needs to be cleared from memory since data is duplicated.
1. Add all set element in list -> al.addAll(s);
2. Sort all the elements in list using -> Collections.sort(al);
public class SortSetProblem {
public static void main(String[] args) {
ArrayList<String> al = new ArrayList();
Set<String> s = new HashSet<>();
s.add("ved");
s.add("prakash");
s.add("sharma");
s.add("apple");
s.add("ved");
s.add("banana");
System.out.println("Before Sorting");
for (String s1 : s) {
System.out.print(" " + s1);
}
System.out.println("After Sorting");
al.addAll(s);
Collections.sort(al);
for (String set : al) {
System.out.print(" " + set);
}
}
}
input - ved prakash sharma apple ved banana
Output - apple banana prakash sharma ved
If you want want the end Collection to be in the form of Set and if you want to define your own natural order rather than that of TreeSet then -
Convert the HashSet into List
Custom sort the List using Comparator
Convert back the List into LinkedHashSet to maintain order
Display the LinkedHashSet
Sample program -
package demo31;
import java.util.*;
public class App26 {
public static void main(String[] args) {
Set<String> set = new HashSet<>();
addElements(set);
List<String> list = new LinkedList<>();
list = convertToList(set);
Collections.sort(list, new Comparator<String>() {
#Override
public int compare(String s1, String s2) {
int flag = s2.length() - s1.length();
if(flag != 0) {
return flag;
} else {
return -s1.compareTo(s2);
}
}
});
Set<String> set2 = new LinkedHashSet<>();
set2 = convertToSet(list);
displayElements(set2);
}
public static void addElements(Set<String> set) {
set.add("Hippopotamus");
set.add("Rhinocerous");
set.add("Zebra");
set.add("Tiger");
set.add("Giraffe");
set.add("Cheetah");
set.add("Wolf");
set.add("Fox");
set.add("Dog");
set.add("Cat");
}
public static List<String> convertToList(Set<String> set) {
List<String> list = new LinkedList<>();
for(String element: set) {
list.add(element);
}
return list;
}
public static Set<String> convertToSet(List<String> list) {
Set<String> set = new LinkedHashSet<>();
for(String element: list) {
set.add(element);
}
return set;
}
public static void displayElements(Set<String> set) {
System.out.println(set);
}
}
Output -
[Hippopotamus, Rhinocerous, Giraffe, Cheetah, Zebra, Tiger, Wolf, Fox, Dog, Cat]
Here the collection has been sorted as -
First - Descending order of String length
Second - Descending order of String alphabetical hierarchy
you can do this in the following ways:
Method 1:
Create a list and store all the hashset values into it
sort the list using Collections.sort()
Store the list back into LinkedHashSet as it preserves the insertion order
Method 2:
Create a treeSet and store all the values into it.
Method 2 is more preferable because the other method consumes lot of time to transfer data back and forth between hashset and list.
We can not decide that the elements of a HashSet would be sorted automatically. But we can sort them by converting into TreeSet or any List like ArrayList or LinkedList etc.
// Create a TreeSet object of class E
TreeSet<E> ts = new TreeSet<E> ();
// Convert your HashSet into TreeSet
ts.addAll(yourHashSet);
System.out.println(ts.toString() + "\t Sorted Automatically");
You can use guava library for the same
Set<String> sortedSet = FluentIterable.from(myHashSet).toSortedSet(new Comparator<String>() {
#Override
public int compare(String s1, String s2) {
// descending order of relevance
//required code
}
});
SortedSet has been added Since java 7
https://docs.oracle.com/javase/8/docs/api/java/util/SortedSet.html
You can wrap it in a TreeSet like this:
Set mySet = new HashSet();
mySet.add(4);
mySet.add(5);
mySet.add(3);
mySet.add(1);
System.out.println("mySet items "+ mySet);
TreeSet treeSet = new TreeSet(mySet);
System.out.println("treeSet items "+ treeSet);
output :
mySet items [1, 3, 4, 5]
treeSet items [1, 3, 4, 5]
Set mySet = new HashSet();
mySet.add("five");
mySet.add("elf");
mySet.add("four");
mySet.add("six");
mySet.add("two");
System.out.println("mySet items "+ mySet);
TreeSet treeSet = new TreeSet(mySet);
System.out.println("treeSet items "+ treeSet);
output:
mySet items [six, four, five, two, elf]
treeSet items [elf, five, four, six, two]
requirement for this method is that the objects of the set/list should be comparable (implement the Comparable interface)
The below is my sample code and its already answered by pointing the code in comments , am still sharing because it contains the complete code
package Collections;
import java.util.*;
public class TestSet {
public static void main(String[] args) {
Set<String> objset = new HashSet<>();
objset.add("test");
objset.add("abc");
objset.add("abc");
objset.add("mas");
objset.add("vas");
Iterator itset = objset.iterator();
while(itset.hasNext())
{
System.out.println(itset.next());
}
TreeSet<String> treeobj = new TreeSet(objset);
System.out.println(treeobj);
}
}
TreeSet treeobj = new TreeSet(objset); here we are invoking the treeset constructor which will call the addAll method to add the objects .
See this below code from the TreeSet class how its mentioned ,
public TreeSet(Collection<? extends E> c) {
this();
addAll(c);
}
Convert HashSet to List then sort it using Collection.sort()
List<String> list = new ArrayList<String>(hset);
Collections.sort(List)
This simple command did the trick for me:
myHashSet.toList.sorted
I used this within a print statement, so if you need to actually persist the ordering, you may need to use TreeSets or other structures proposed on this thread.
I am storing an arrayList as my keys in a TreeMap but I am getting this exception
java.lang.ClassCastException: java.util.ArrayList cannot be cast to java.lang.Comparable
I copy the content of array to ArrayList and try to store that arrayList as my keys in the Map
My Code is :
TreeMap< ArrayList<Integer> , Integer > bandsMap = new TreeMap< ArrayList<Integer> , Integer >();
ArrayList< Integer > erfcn = new ArrayList< Integer >();
for (int index = 0; index < frequencies.length; index++)
erfcn.add(frequencies[index]);
bandsMap.put( erfcn , band_number);
for (Integer value : bandsMap.values()) {
System.out.println("Value = " + value + "\n");
}
Any Idea ?
Thanks
A tree map maintains its keys in sorted order. The ArrayList class does not define any ordering, so it cannot be used as a key directly. You can supply an external comparator to impose an order though, but you must define an ordering that makes sense to you:
TreeMap<ArrayList<Integer>, Integer> bandsMap = new TreeMap<>(
new Comparator<ArrayList<Integer>>() {
public int compare(ArrayList<Integer> lst1, ArrayList<Integer> lst2) {
// return 1 if lst1 > lst2, 0 if equal, -1 if lst1 < lst2
}
});
Alternatively, if you don't have to maintain the keys in any particular order, use a HashMap instead.
ArrayLists don't implement Comparable, so you need to use an unsorted map, like HashMap, or tell the TreeMap how to sort your ArrayLists by using this constructor.
The error itself indicates the problem. ArrayList class does not implement java.lang.Comparable interface and the TreeMap is expecting the key to implement comparable interface. Hence it is causing the exception.
Refer ArrayList documentation.
As we cannot go modify ArrayList, you can use external comparator to make ArrayList work as a key to the TreeMap. You just need to override compare() method in it.
You can't compare two lists, you have to either change list to some other structure, or make your own list using Comparable interface. Good solution is also to wrap up list in new class which implements Comparable and implement just this method from interface.
Check out this
public class Fruit implements Comparable<Fruit>{
public int compareTo(Fruit compareFruit) {
//your code here
}
}
and this link. Hope it helps.
If you really want to use ArrayList as key in TreeMap, then you need to write Comparator for it and pass is using constructor
Using List as key in Tree is not a good idea, please review your design.
Given what the other answers have said about Lists not implementing Comparable, you could create your own class to act as TreeMap keys, extending ArrayList and implementing Comparable:
class KeyList extends ArrayList<Integer> implements Comparable<ArrayList<Integer>> {
public int compareTo(ArrayList<Integer> list) {
//decide how to compare ArrayLists, then implement it here
return 0;
}
}
Then you can create your TreeMap:
new TreeMap<KeyList, Integer>();
I can not initialize a List as in the following code:
List<String> supplierNames = new List<String>();
supplierNames.add("sup1");
supplierNames.add("sup2");
supplierNames.add("sup3");
System.out.println(supplierNames.get(1));
I face the following error:
Cannot instantiate the type List<String>
How can I instantiate List<String>?
If you check the API for List you'll notice it says:
Interface List<E>
Being an interface means it cannot be instantiated (no new List() is possible).
If you check that link, you'll find some classes that implement List:
All Known Implementing Classes:
AbstractList, AbstractSequentialList, ArrayList, AttributeList, CopyOnWriteArrayList, LinkedList, RoleList, RoleUnresolvedList, Stack, Vector
Some of those can be instantiated (the ones that are not defined as abstract class). Use their links to know more about them, I.E: to know which fits better your needs.
The 3 most commonly used ones probably are:
List<String> supplierNames1 = new ArrayList<String>();
List<String> supplierNames2 = new LinkedList<String>();
List<String> supplierNames3 = new Vector<String>();
Bonus:
You can also instantiate it with values, in an easier way, using the Arrays class, as follows:
List<String> supplierNames = Arrays.asList("sup1", "sup2", "sup3");
System.out.println(supplierNames.get(1));
But note you are not allowed to add more elements to that list, as it's fixed-size.
Can't instantiate an interface but there are few implementations:
JDK2
List<String> list = Arrays.asList("one", "two", "three");
JDK7
//diamond operator
List<String> list = new ArrayList<>();
list.add("one");
list.add("two");
list.add("three");
JDK8
List<String> list = Stream.of("one", "two", "three").collect(Collectors.toList());
JDK9
// creates immutable lists, so you can't modify such list
List<String> immutableList = List.of("one", "two", "three");
// if we want mutable list we can copy content of immutable list
// to mutable one for instance via copy-constructor (which creates shallow copy)
List<String> mutableList = new ArrayList<>(List.of("one", "two", "three"));
Plus there are lots of other ways supplied by other libraries like Guava.
List<String> list = Lists.newArrayList("one", "two", "three");
List is an Interface, you cannot instantiate an Interface, because interface is a convention, what methods should have your classes. In order to instantiate, you need some realizations(implementations) of that interface. Try the below code with very popular implementations of List interface:
List<String> supplierNames = new ArrayList<String>();
or
List<String> supplierNames = new LinkedList<String>();
You will need to use ArrayList<String> or such.
List<String> is an interface.
Use this:
import java.util.ArrayList;
...
List<String> supplierNames = new ArrayList<String>();
List is an interface, and you can not initialize an interface. Instantiate an implementing class instead.
Like:
List<String> abc = new ArrayList<String>();
List<String> xyz = new LinkedList<String>();
In most cases you want simple ArrayList - an implementation of List
Before JDK version 7
List<String> list = new ArrayList<String>();
JDK 7 and later you can use the diamond operator
List<String> list = new ArrayList<>();
Further informations are written here Oracle documentation - Collections
List is just an interface, a definition of some generic list. You need to provide an implementation of this list interface. Two most common are:
ArrayList - a list implemented over an array
List<String> supplierNames = new ArrayList<String>();
LinkedList - a list implemented like an interconnected chain of elements
List<String> supplierNames = new LinkedList<String>();
Depending on what kind of List you want to use, something like
List<String> supplierNames = new ArrayList<String>();
should get you going.
List is the interface, ArrayList is one implementation of the List interface. More implementations that may better suit your needs can be found by reading the JavaDocs of the List interface.
If you just want to create an immutable List<T> with only one object in it, you can use this API:
List<String> oneObjectList = Collections.singletonList("theOnlyObject”);
More info: docs
List is an Interface . You cant use List to initialize it.
List<String> supplierNames = new ArrayList<String>();
These are the some of List impelemented classes,
ArrayList, LinkedList, Vector
You could use any of this as per your requirement. These each classes have its own features.
Just in case, any one still lingering around this question. Because, i see one or two new users again asking the same question and everyone telling then , No you can't do that, Dear Prudence, Apart from all the answers given here, I would like to provide additional Information -
Yes you can actually do, List list = new List();
But at the cost of writing implementations of all the methods of Interfaces.
The notion is not simply List list = new List(); but
List<Integer> list = new List<Integer>(){
#Override
public int size() {
// TODO Auto-generated method stub
return 0;
}
#Override
public boolean isEmpty() {
// TODO Auto-generated method stub
return false;
}
#Override
public boolean contains(Object o) {
// TODO Auto-generated method stub
return false;
}
..... and So on (Cant write all methods.)
This is an example of Anonymous class. Its correct when someone states , No you cant instantiate an interface, and that's right. But you can never say , You CANT write List list = new List(); but, evidently you can do that and that's a hard statement to make that You can't do.
Instead of :
List<String> supplierNames = new List<String>();
Write this if you are using latest JDK:
List<String> supplierNames = new ArrayList<>();
It's the correct way of initializing a List.
We created soyuz-to to simplify 1 problem: how to convert X to Y (e.g. String to Integer). Constructing of an object is also kind of conversion so it has a simple function to construct Map, List, Set:
import io.thedocs.soyuz.to;
List<String> names = to.list("John", "Fedor");
Please check it - it has a lot of other useful features
What is the syntax for making a List of arrays in Java?
I have tried the following:
List<int[]> A = new List<int[]>();
and a lot of other things.
I need to be able to reorder the int arrays, but the elements of the int arrays need not to be changed. If this is not possible, why?
Thank you.
Firstly, you can't do new List(); it is an interface.
To make a list of int Arrays, do something like this :
List<int[]> myList = new ArrayList<int[]>();
P.S. As per the comment, package for List is java.util.List and for ArrayList java.util.ArrayList
List<Integer[]> integerList = new ArrayList<Integer[]>();
Use the object instead of the primitive, unless this is before Java 1.5 as it handles the autoboxing automatically.
As far as the sorting goes:
Collections.sort(integerList); //Sort the entire List
and for each array (probably what you want)
for(Integer[] currentArray : integerList)
{
Arrays.sort(currentArray);
}
List is an interface, not a class. You have to choose what kind of list. In most cases an ArrayList is chosen.
List a = new ArrayList();
You've mentioned that you want to store an int array in it, so you can specify the type that a list contains.
List<int[]> a = new ArrayList<int[]>();
While you can have a collection (such as a list) of "int[]", you cannot have a collection of "int". This is because arrays are objects, but an "int" is a primitive.
Can somebody please explain me why I can't cast List<> to ArrayList<> with first approach and I do with second one? Thank you.
First approach:
ArrayList<Task> tmp = ((ArrayList<Task>)mTrackytAdapter.getAllTasks(token));
Second approach:
ArrayList<Task> tmp = new ArrayList<Task>(mTrackytAdapter.getAllTasks(token));
When you do the second one, you're making a new arraylist, you're not trying to pretend the other list is an arraylist.
I mean, what if the original list is implemented as a linkedlist, or some custom list? You won't know. The second approach is preferred if you really need to make an arraylist from the result. But you can just leave it as a list, that's one of the best advantages of using Interfaces!
When you are using second approach you are initializing arraylist with its predefined values.
Like generally we do
**ArrayList listofStrings = new ArrayList<>();
**
Let's say you have an array with values, now you want to convert this array into arraylist.
you need to first get the list from the array using Arrays utils.
Because the ArrayList is concrete type that implement List interface. It is not guaranteed that method asList, will return this type of implementation.
List<String> listofOptions = (List<String>) Arrays.asList(options);
then you can user constructoru of an arraylist to instantiate with predefined values.
ArrayList<String> arrlistofOptions = new ArrayList<String>(list);
So your second approach is working that you have passed values which will intantiate arraylist with the list elements.
More over
ArrayList that is returned from Arrays.asList is not an actual arraylist, it is just a wrapper which doesnt allows any modification in the list.
If you try to add or remove over Arrays.asList it will give you
UnsupportedOperationException
Try running the following code:
List<String> listOfString = Arrays.asList("Hello", "World");
ArrayList<String> arrayListOfString = new ArrayList(listOfString);
System.out.println(listOfString.getClass());
System.out.println(arrayListOfString.getClass());
You'll get the following result:
class java.util.Arrays$ArrayList
class java.util.ArrayList
So, that means they're 2 different classes that aren't extending each other. java.util.Arrays$ArrayList signifies the private class named ArrayList (inner class of Arrays class) and java.util.ArrayList signifies the public class named ArrayList. Thus, casting from java.util.Arrays$ArrayList to java.util.ArrayList and vice versa are irrelevant/not available.
The second approach is clearly wrong if you want to cast. It instantiate a new ArrayList.
However the first approach should work just fine, if and only if getAllTasks return an ArrayList.
It is really needed for you to have an ArrayList ? isn't the List interface enough ? What you are doing can leads to Runtime Exception if the type isn't correct.
If getAllTasks() return an ArrayList you should change the return type in the class definition and then you won't need a cast and if it's returning something else, you can't cast to ArrayList.
Just try this :
ArrayList<SomeClass> arrayList;
public SomeConstructor(List<SomeClass> listData) {
arrayList.addAll(listData);
}
You can cast List<> to ArrayList<> if you understand what you doing. Java compiler won't block it.
But:
It's bad practice to casting parent type to child type (or interface to implementation type) without checking.
This way better:
if (list instanceof ArrayList<Task>) {
ArrayList<Task> arraylist = (ArrayList<Task>) list;
}
Maybe you don't need implementation type as reference. Look SonarQube warning https://sbforge.org/sonar/rules/show/squid:S1319. You can avoid this casting in the most cases.
You can use Guava method:
ArrayList<Task> arraylist = Lists.newArrayList(list);
The first approach is trying to cast the list but this would work only if the List<> were an ArrayList<>. That is not the case. So you need the second approach, that is building a new ArrayList<> with the elements of the List<>
Because in the first one , you're trying to convert a collection to an ArrayList.
In the 2nd one , you just use the built in constructor of ArrayList
May be:
ArrayList<ServiceModel> services = new ArrayList<>(parking.getServices());
intent.putExtra("servicios",services);
import java.util.List;
import java.util.Arrays;
import java.util.*;
public class Merge
{
public static void main(String[] args) {
// This is normal way
// List<Integer> l1 = new ArrayList<Integer>(); l1.add(2); l1.add(5); l1.add(10); l1.add(22);
// List<Integer> l2 = new ArrayList<Integer>(); l2.add(3); l2.add(8); l2.add(15);
//Array.asList only have the list interface, but ArrayList is inherited from List Interface with few more property like ArrayList.remove()
List<Integer> templ1 = Arrays.asList(2,5,10,22);
List<Integer> templ2 = Arrays.asList(3,8,12);
//So creation of ArrayList with the given list is required, then only ArrayList.remove function works.
List<Integer> l1 = new ArrayList<Integer>(templ1);
List<Integer> l2 = new ArrayList<Integer>(templ2);
List<Integer> l3 = new ArrayList<Integer>();
Iterator itr1 = l1.iterator();
while(itr1.hasNext()){
int x = (Integer) itr1.next();
Iterator itr2 = l2.iterator();
while(itr2.hasNext()) {
int y = (Integer) itr2.next();
if(x < y) {
l3.add(x);
break;
}
else{
l3.add(y);
itr2.remove();
}
}
}
Iterator it = l1.iterator();
while (it.hasNext()){
int k = (Integer) it.next();
if (l3.contains(k)){
continue;
}
else{
l3.add(k);
System.out.println(k);
}
}
Iterator itr2 = l2.iterator();
while (itr2.hasNext()){
int k = (Integer) itr2.next();
l3.add(k);
}
System.out.println(l3);
}
}