Related
So I need to choose between
HashMap<String[], Object>
HashMap<ArrayList<String>,Object>
My input Parameter is: ArrayList<String> in.
The whole ArrayList<String> in cannot be the key, since it does contain elements, which are not supposed to be like a Primary Key in a database. I do know, that the first n elements of the incoming ArrayList<String> in supposed to be the primary Keys.
Which one would be faster?
Scenario:
HashMap<ArrayList<String>, Object> hmAL = new HashMap<>();
HashMap<String[], Object> hmSA = new HashMap<>();
ArrayList<String> in = new ArrayList<>();
fillWithStuff(in);
//Which one would be faster?
getObject(in,hmAL,5);
getObject(in,hmSA,5);
With Option 1:
private Object getObject(ArrayList<String> in, HashMap<ArrayList<String>, Object> hm, int n){
return hm.get(in.sublist(0,n));
}
With Option 2:
private Object getObject(ArrayList<String> in, HashMap<String[], Object> hm, int n){
String[] temp = new String[n];
for(int i=0; i<n; i++)
temp[i]=in.get(i);
return hm.get(temp);
}
Considering:
Which is faster? Short the list, or copy to an array?
I'm wondering, which hash (since it is a HashMap) would be faster. Hashing of a ArrayList, or an equal-sized array. Or doesn't it make any difference?
Using String[] is not a good idea because it does not implement hashCode(). This means if you have 2 string arrays which are different objects but with the exact same values, the map will not find it.
The implementation of 'hashCode` seems to use each of the string elements hashcode so the lookup in a map would succeed. So I'd go with this one.
That said, I would rather build a key myself based on the objects in the list.
Dealing with copying only
The subList method is implemented very efficiently in Java 7+, not requiring any copying at all. It simply returns a view directly onto the original array. Thus, in Java 7+, it will be faster than the copy element by element method. However, in Java 6, both ways are essentially equivalent.
Dealing with the method as a whole
If you look at the whole method, your choice is no longer a choice. If you want the method to function, you will have to use the first implementation. Array hashCode() does not look at the elements inside it---only the identity of the array. Because you are creating the array in your method, the Map.get() will necessary return null.
On the other hand, the List.hashCode() method runs a hash on all of the contained elements, meaning that it will successfully match if all of the contained elements are the same.
Your choice is clear.
Just to add on above two answers, I have tested in Java 7 and found on an average with list it's 50 times faster with 2000000 total elements and 1000000 elements which participate in calculating hashcode i.e. primary keys (hypothetical number). Below is the program.
public class TestHashing {
public static void main(String[] args) {
HashMap<ArrayList<String>, Object> hmAL = new HashMap();
HashMap<String[], Object> hmSA = new HashMap<>();
ArrayList<String> in = new ArrayList<>();
fillWithStuff(in);
// Which one would be faster?
long start = System.nanoTime();
getObject(in, hmAL, 1000000);
long end = System.nanoTime();
long firstTime = (end-start);
System.out.println("firstTime :: "+ firstTime);
start = System.nanoTime();
getObject1(in, hmSA, 1000000);
end = System.nanoTime();
long secondTime = (end-start);
System.out.println("secondTime :: "+ secondTime);
System.out.println("First is faster by "+ secondTime/firstTime);
}
private static void fillWithStuff(ArrayList<String> in) {
for(int i =0; i< 2000000; i++) {
in.add(i+"");
}
}
private static Object getObject(ArrayList<String> in,
HashMap<ArrayList<String>, Object> hm, int n) {
return hm.get(in.subList(0, n));
}
private static Object getObject1(ArrayList<String> in, HashMap<String[], Object> hm, int n){
String[] temp = new String[n];
for(int i=0; i<n; i++)
temp[i]=in.get(i);
return hm.get(temp);
}
}
Output
firstTime :: 218000
secondTime :: 11627000
First is faster by 53
I have one Arraylist of String and I have added Some Duplicate Value in that. and i just wanna remove that Duplicate value So how to remove it.
Here Example I got one Idea.
List<String> list = new ArrayList<String>();
list.add("Krishna");
list.add("Krishna");
list.add("Kishan");
list.add("Krishn");
list.add("Aryan");
list.add("Harm");
System.out.println("List"+list);
for (int i = 1; i < list.size(); i++) {
String a1 = list.get(i);
String a2 = list.get(i-1);
if (a1.equals(a2)) {
list.remove(a1);
}
}
System.out.println("List after short"+list);
But is there any Sufficient way remove that Duplicate form list. with out using For loop ?
And ya i can do it by using HashSet or some other way but using array list only.
would like to have your suggestion for that. thank you for your answer in advance.
You can create a LinkedHashSet from the list. The LinkedHashSet will contain each element only once, and in the same order as the List. Then create a new List from this LinkedHashSet. So effectively, it's a one-liner:
list = new ArrayList<String>(new LinkedHashSet<String>(list))
Any approach that involves List#contains or List#remove will probably decrease the asymptotic running time from O(n) (as in the above example) to O(n^2).
EDIT For the requirement mentioned in the comment: If you want to remove duplicate elements, but consider the Strings as equal ignoring the case, then you could do something like this:
Set<String> toRetain = new TreeSet<String>(String.CASE_INSENSITIVE_ORDER);
toRetain.addAll(list);
Set<String> set = new LinkedHashSet<String>(list);
set.retainAll(new LinkedHashSet<String>(toRetain));
list = new ArrayList<String>(set);
It will have a running time of O(n*logn), which is still better than many other options. Note that this looks a little bit more complicated than it might have to be: I assumed that the order of the elements in the list may not be changed. If the order of the elements in the list does not matter, you can simply do
Set<String> set = new TreeSet<String>(String.CASE_INSENSITIVE_ORDER);
set.addAll(list);
list = new ArrayList<String>(set);
if you want to use only arraylist then I am worried there is no better way which will create a huge performance benefit. But by only using arraylist i would check before adding into the list like following
void addToList(String s){
if(!yourList.contains(s))
yourList.add(s);
}
In this cases using a Set is suitable.
You can make use of Google Guava utilities, as shown below
list = ImmutableSet.copyOf(list).asList();
This is probably the most efficient way of eliminating the duplicates from the list and interestingly, it preserves the iteration order as well.
UPDATE
But, in case, you don't want to involve Guava then duplicates can be removed as shown below.
ArrayList<String> list = new ArrayList<String>();
list.add("Krishna");
list.add("Krishna");
list.add("Kishan");
list.add("Krishn");
list.add("Aryan");
list.add("Harm");
System.out.println("List"+list);
HashSet hs = new HashSet();
hs.addAll(list);
list.clear();
list.addAll(hs);
But, of course, this will destroys the iteration order of the elements in the ArrayList.
Shishir
Java 8 stream function
You could use the distinct function like above to get the distinct elements of the list,
stringList.stream().distinct();
From the documentation,
Returns a stream consisting of the distinct elements (according to Object.equals(Object)) of this stream.
Another way, if you do not wish to use the equals method is by using the collect function like this,
stringList.stream()
.collect(Collectors.toCollection(() ->
new TreeSet<String>((p1, p2) -> p1.compareTo(p2))
));
From the documentation,
Performs a mutable reduction operation on the elements of this stream using a Collector.
Hope that helps.
Simple function for removing duplicates from list
private void removeDuplicates(List<?> list)
{
int count = list.size();
for (int i = 0; i < count; i++)
{
for (int j = i + 1; j < count; j++)
{
if (list.get(i).equals(list.get(j)))
{
list.remove(j--);
count--;
}
}
}
}
Example:
Input: [1, 2, 2, 3, 1, 3, 3, 2, 3, 1, 2, 3, 3, 4, 4, 4, 1]
Output: [1, 2, 3, 4]
List<String> list = new ArrayList<String>();
list.add("Krishna");
list.add("Krishna");
list.add("Kishan");
list.add("Krishn");
list.add("Aryan");
list.add("Harm");
HashSet<String> hs=new HashSet<>(list);
System.out.println("=========With Duplicate Element========");
System.out.println(list);
System.out.println("=========Removed Duplicate Element========");
System.out.println(hs);
I don't think the list = new ArrayList<String>(new LinkedHashSet<String>(list)) is not the best way , since we are using the LinkedHashset(We could use directly LinkedHashset instead of ArrayList),
Solution:
import java.util.ArrayList;
public class Arrays extends ArrayList{
#Override
public boolean add(Object e) {
if(!contains(e)){
return super.add(e);
}else{
return false;
}
}
public static void main(String[] args) {
Arrays element=new Arrays();
element.add(1);
element.add(2);
element.add(2);
element.add(3);
System.out.println(element);
}
}
Output:
[1, 2, 3]
Here I am extending the ArrayList , as I am using the it with some changes by overriding the add method.
public List<Contact> removeDuplicates(List<Contact> list) {
// Set set1 = new LinkedHashSet(list);
Set set = new TreeSet(new Comparator() {
#Override
public int compare(Object o1, Object o2) {
if(((Contact)o1).getId().equalsIgnoreCase(((Contact)2).getId()) ) {
return 0;
}
return 1;
}
});
set.addAll(list);
final List newList = new ArrayList(set);
return newList;
}
This will be the best way
List<String> list = new ArrayList<String>();
list.add("Krishna");
list.add("Krishna");
list.add("Kishan");
list.add("Krishn");
list.add("Aryan");
list.add("Harm");
Set<String> set=new HashSet<>(list);
It is better to use HastSet
1-a) A HashSet holds a set of objects, but in a way that it allows you to easily and quickly determine whether an object is already in the set or not. It does so by internally managing an array and storing the object using an index which is calculated from the hashcode of the object. Take a look here
1-b) HashSet is an unordered collection containing unique elements. It has the standard collection operations Add, Remove, Contains, but since it uses a hash-based implementation, these operation are O(1). (As opposed to List for example, which is O(n) for Contains and Remove.) HashSet also provides standard set operations such as union, intersection, and symmetric difference.Take a look here
2) There are different implementations of Sets. Some make insertion and lookup operations super fast by hashing elements. However that means that the order in which the elements were added is lost. Other implementations preserve the added order at the cost of slower running times.
The HashSet class in C# goes for the first approach, thus not preserving the order of elements. It is much faster than a regular List. Some basic benchmarks showed that HashSet is decently faster when dealing with primary types (int, double, bool, etc.). It is a lot faster when working with class objects. So that point is that HashSet is fast.
The only catch of HashSet is that there is no access by indices. To access elements you can either use an enumerator or use the built-in function to convert the HashSet into a List and iterate through that.Take a look here
Without a loop, No! Since ArrayList is indexed by order rather than by key, you can not found the target element without iterate the whole list.
A good practice of programming is to choose proper data structure to suit your scenario. So if Set suits your scenario the most, the discussion of implementing it with List and trying to find the fastest way of using an improper data structure makes no sense.
public static void main(String[] args) {
#SuppressWarnings("serial")
List<Object> lst = new ArrayList<Object>() {
#Override
public boolean add(Object e) {
if(!contains(e))
return super.add(e);
else
return false;
}
};
lst.add("ABC");
lst.add("ABC");
lst.add("ABCD");
lst.add("ABCD");
lst.add("ABCE");
System.out.println(lst);
}
This is the better way
list = list.stream().distinct().collect(Collectors.toList());
This could be one of the solutions using Java8 Stream API. Hope this helps.
public void removeDuplicates() {
ArrayList<Object> al = new ArrayList<Object>();
al.add("java");
al.add('a');
al.add('b');
al.add('a');
al.add("java");
al.add(10.3);
al.add('c');
al.add(14);
al.add("java");
al.add(12);
System.out.println("Before Remove Duplicate elements:" + al);
for (int i = 0; i < al.size(); i++) {
for (int j = i + 1; j < al.size(); j++) {
if (al.get(i).equals(al.get(j))) {
al.remove(j);
j--;
}
}
}
System.out.println("After Removing duplicate elements:" + al);
}
Before Remove Duplicate elements:
[java, a, b, a, java, 10.3, c, 14, java, 12]
After Removing duplicate elements:
[java, a, b, 10.3, c, 14, 12]
Using java 8:
public static <T> List<T> removeDuplicates(List<T> list) {
return list.stream().collect(Collectors.toSet()).stream().collect(Collectors.toList());
}
In case you just need to remove the duplicates using only ArrayList, no other Collection classes, then:-
//list is the original arraylist containing the duplicates as well
List<String> uniqueList = new ArrayList<String>();
for(int i=0;i<list.size();i++) {
if(!uniqueList.contains(list.get(i)))
uniqueList.add(list.get(i));
}
Hope this helps!
private static void removeDuplicates(List<Integer> list)
{
Collections.sort(list);
int count = list.size();
for (int i = 0; i < count; i++)
{
if(i+1<count && list.get(i)==list.get(i+1)){
list.remove(i);
i--;
count--;
}
}
}
public static List<String> removeDuplicateElements(List<String> array){
List<String> temp = new ArrayList<String>();
List<Integer> count = new ArrayList<Integer>();
for (int i=0; i<array.size()-2; i++){
for (int j=i+1;j<array.size()-1;j++)
{
if (array.get(i).compareTo(array.get(j))==0) {
count.add(i);
int kk = i;
}
}
}
for (int i = count.size()+1;i>0;i--) {
array.remove(i);
}
return array;
}
}
Is there a "computationally" quick way to get the count of an iterator?
int i = 0;
for ( ; some_iterator.hasNext() ; ++i ) some_iterator.next();
... seems like a waste of CPU cycles.
Using Guava library:
int size = Iterators.size(iterator);
Internally it just iterates over all elements so its just for convenience.
If you've just got the iterator then that's what you'll have to do - it doesn't know how many items it's got left to iterate over, so you can't query it for that result. There are utility methods that will seem to do this efficiently (such as Iterators.size() in Guava), but underneath they're just consuming the iterator and counting as they go, the same as in your example.
However, many iterators come from collections, which you can often query for their size. And if it's a user made class you're getting the iterator for, you could look to provide a size() method on that class.
In short, in the situation where you only have the iterator then there's no better way, but much more often than not you have access to the underlying collection or object from which you may be able to get the size directly.
Your code will give you an exception when you reach the end of the iterator. You could do:
int i = 0;
while(iterator.hasNext()) {
i++;
iterator.next();
}
If you had access to the underlying collection, you would be able to call coll.size()...
EDIT
OK you have amended...
You will always have to iterate. Yet you can use Java 8, 9 to do the counting without looping explicitely:
Iterable<Integer> newIterable = () -> iter;
long count = StreamSupport.stream(newIterable.spliterator(), false).count();
Here is a test:
public static void main(String[] args) throws IOException {
Iterator<Integer> iter = Arrays.asList(1, 2, 3, 4, 5).iterator();
Iterable<Integer> newIterable = () -> iter;
long count = StreamSupport.stream(newIterable.spliterator(), false).count();
System.out.println(count);
}
This prints:
5
Interesting enough you can parallelize the count operation here by changing the parallel flag on this call:
long count = StreamSupport.stream(newIterable.spliterator(), *true*).count();
Using Guava library, another option is to convert the Iterable to a List.
List list = Lists.newArrayList(some_iterator);
int count = list.size();
Use this if you need also to access the elements of the iterator after getting its size. By using Iterators.size() you no longer can access the iterated elements.
If all you have is the iterator, then no, there is no "better" way. If the iterator comes from a collection you could as that for size.
Keep in mind that Iterator is just an interface for traversing distinct values, you would very well have code such as this
new Iterator<Long>() {
final Random r = new Random();
#Override
public boolean hasNext() {
return true;
}
#Override
public Long next() {
return r.nextLong();
}
#Override
public void remove() {
throw new IllegalArgumentException("Not implemented");
}
};
or
new Iterator<BigInteger>() {
BigInteger next = BigInteger.ZERO;
#Override
public boolean hasNext() {
return true;
}
#Override
public BigInteger next() {
BigInteger current = next;
next = next.add(BigInteger.ONE);
return current;
}
#Override
public void remove() {
throw new IllegalArgumentException("Not implemented");
}
};
There is no more efficient way, if all you have is the iterator. And if the iterator can only be used once, then getting the count before you get the iterator's contents is ... problematic.
The solution is either to change your application so that it doesn't need the count, or to obtain the count by some other means. (For example, pass a Collection rather than Iterator ...)
for Java 8 you could use,
public static int getIteratorSize(Iterator iterator){
AtomicInteger count = new AtomicInteger(0);
iterator.forEachRemaining(element -> {
count.incrementAndGet();
});
return count.get();
}
To get the size of an Iterable
Iterable<Users> users = usersRepository.findUsersByLocation("IND");
Now assert the size of users of Type Iterable
assertEquals(2, ((Collection<Users>)users).size());
To get quickly the size :
[...iterator].length
iterator object contains the same number of elements what your collection contained.
List<E> a =...;
Iterator<E> i = a.iterator();
int size = a.size();//Because iterators size is equal to list a's size.
But instead of getting the size of iterator and iterating through index 0 to that size, it is better to iterate through the method next() of the iterator.
I have an ArrayList that I want to iterate over. While iterating over it I have to remove elements at the same time. Obviously this throws a java.util.ConcurrentModificationException.
What is the best practice to handle this problem? Should I clone the list first?
I remove the elements not in the loop itself but another part of the code.
My code looks like this:
public class Test() {
private ArrayList<A> abc = new ArrayList<A>();
public void doStuff() {
for (A a : abc)
a.doSomething();
}
public void removeA(A a) {
abc.remove(a);
}
}
a.doSomething might call Test.removeA();
Two options:
Create a list of values you wish to remove, adding to that list within the loop, then call originalList.removeAll(valuesToRemove) at the end
Use the remove() method on the iterator itself. Note that this means you can't use the enhanced for loop.
As an example of the second option, removing any strings with a length greater than 5 from a list:
List<String> list = new ArrayList<String>();
...
for (Iterator<String> iterator = list.iterator(); iterator.hasNext(); ) {
String value = iterator.next();
if (value.length() > 5) {
iterator.remove();
}
}
From the JavaDocs of the ArrayList
The iterators returned by this class's iterator and listIterator
methods are fail-fast: if the list is structurally modified at any
time after the iterator is created, in any way except through the
iterator's own remove or add methods, the iterator will throw a
ConcurrentModificationException.
You are trying to remove value from list in advanced "for loop", which is not possible, even if you apply any trick (which you did in your code).
Better way is to code iterator level as other advised here.
I wonder how people have not suggested traditional for loop approach.
for( int i = 0; i < lStringList.size(); i++ )
{
String lValue = lStringList.get( i );
if(lValue.equals("_Not_Required"))
{
lStringList.remove(lValue);
i--;
}
}
This works as well.
In Java 8 you can use the Collection Interface and do this by calling the removeIf method:
yourList.removeIf((A a) -> a.value == 2);
More information can be found here
You should really just iterate back the array in the traditional way
Every time you remove an element from the list, the elements after will be push forward. As long as you don't change elements other than the iterating one, the following code should work.
public class Test(){
private ArrayList<A> abc = new ArrayList<A>();
public void doStuff(){
for(int i = (abc.size() - 1); i >= 0; i--)
abc.get(i).doSomething();
}
public void removeA(A a){
abc.remove(a);
}
}
While iterating the list, if you want to remove the element is possible. Let see below my examples,
ArrayList<String> names = new ArrayList<String>();
names.add("abc");
names.add("def");
names.add("ghi");
names.add("xyz");
I have the above names of Array list. And i want to remove the "def" name from the above list,
for(String name : names){
if(name.equals("def")){
names.remove("def");
}
}
The above code throws the ConcurrentModificationException exception because you are modifying the list while iterating.
So, to remove the "def" name from Arraylist by doing this way,
Iterator<String> itr = names.iterator();
while(itr.hasNext()){
String name = itr.next();
if(name.equals("def")){
itr.remove();
}
}
The above code, through iterator we can remove the "def" name from the Arraylist and try to print the array, you would be see the below output.
Output : [abc, ghi, xyz]
Do the loop in the normal way, the java.util.ConcurrentModificationException is an error related to the elements that are accessed.
So try:
for(int i = 0; i < list.size(); i++){
lista.get(i).action();
}
Here is an example where I use a different list to add the objects for removal, then afterwards I use stream.foreach to remove elements from original list :
private ObservableList<CustomerTableEntry> customersTableViewItems = FXCollections.observableArrayList();
...
private void removeOutdatedRowsElementsFromCustomerView()
{
ObjectProperty<TimeStamp> currentTimestamp = new SimpleObjectProperty<>(TimeStamp.getCurrentTime());
long diff;
long diffSeconds;
List<Object> objectsToRemove = new ArrayList<>();
for(CustomerTableEntry item: customersTableViewItems) {
diff = currentTimestamp.getValue().getTime() - item.timestamp.getValue().getTime();
diffSeconds = diff / 1000 % 60;
if(diffSeconds > 10) {
// Element has been idle for too long, meaning no communication, hence remove it
System.out.printf("- Idle element [%s] - will be removed\n", item.getUserName());
objectsToRemove.add(item);
}
}
objectsToRemove.stream().forEach(o -> customersTableViewItems.remove(o));
}
One option is to modify the removeA method to this -
public void removeA(A a,Iterator<A> iterator) {
iterator.remove(a);
}
But this would mean your doSomething() should be able to pass the iterator to the remove method. Not a very good idea.
Can you do this in two step approach :
In the first loop when you iterate over the list , instead of removing the selected elements , mark them as to be deleted. For this , you may simply copy these elements ( shallow copy ) into another List.
Then , once your iteration is done , simply do a removeAll from the first list all elements in the second list.
In my case, the accepted answer is not working, It stops Exception but it causes some inconsistency in my List. The following solution is perfectly working for me.
List<String> list = new ArrayList<>();
List<String> itemsToRemove = new ArrayList<>();
for (String value: list) {
if (value.length() > 5) { // your condition
itemsToRemove.add(value);
}
}
list.removeAll(itemsToRemove);
In this code, I have added the items to remove, in another list and then used list.removeAll method to remove all required items.
Instead of using For each loop, use normal for loop. for example,the below code removes all the element in the array list without giving java.util.ConcurrentModificationException. You can modify the condition in the loop according to your use case.
for(int i=0; i<abc.size(); i++) {
e.remove(i);
}
Sometimes old school is best. Just go for a simple for loop but make sure you start at the end of the list otherwise as you remove items you will get out of sync with your index.
List<String> list = new ArrayList<>();
for (int i = list.size() - 1; i >= 0; i--) {
if ("removeMe".equals(list.get(i))) {
list.remove(i);
}
}
You can also use CopyOnWriteArrayList instead of an ArrayList. This is the latest recommended approach by from JDK 1.5 onwards.
Do somehting simple like this:
for (Object object: (ArrayList<String>) list.clone()) {
list.remove(object);
}
An alternative Java 8 solution using stream:
theList = theList.stream()
.filter(element -> !shouldBeRemoved(element))
.collect(Collectors.toList());
In Java 7 you can use Guava instead:
theList = FluentIterable.from(theList)
.filter(new Predicate<String>() {
#Override
public boolean apply(String element) {
return !shouldBeRemoved(element);
}
})
.toImmutableList();
Note, that the Guava example results in an immutable list which may or may not be what you want.
for (A a : new ArrayList<>(abc)) {
a.doSomething();
abc.remove(a);
}
"Should I clone the list first?"
That will be the easiest solution, remove from the clone, and copy the clone back after removal.
An example from my rummikub game:
SuppressWarnings("unchecked")
public void removeStones() {
ArrayList<Stone> clone = (ArrayList<Stone>) stones.clone();
// remove the stones moved to the table
for (Stone stone : stones) {
if (stone.isOnTable()) {
clone.remove(stone);
}
}
stones = (ArrayList<Stone>) clone.clone();
sortStones();
}
I arrive late I know but I answer this because I think this solution is simple and elegant:
List<String> listFixed = new ArrayList<String>();
List<String> dynamicList = new ArrayList<String>();
public void fillingList() {
listFixed.add("Andrea");
listFixed.add("Susana");
listFixed.add("Oscar");
listFixed.add("Valeria");
listFixed.add("Kathy");
listFixed.add("Laura");
listFixed.add("Ana");
listFixed.add("Becker");
listFixed.add("Abraham");
dynamicList.addAll(listFixed);
}
public void updatingListFixed() {
for (String newList : dynamicList) {
if (!listFixed.contains(newList)) {
listFixed.add(newList);
}
}
//this is for add elements if you want eraser also
String removeRegister="";
for (String fixedList : listFixed) {
if (!dynamicList.contains(fixedList)) {
removeResgister = fixedList;
}
}
fixedList.remove(removeRegister);
}
All this is for updating from one list to other and you can make all from just one list
and in method updating you check both list and can eraser or add elements betwen list.
This means both list always it same size
Use Iterator instead of Array List
Have a set be converted to iterator with type match
And move to the next element and remove
Iterator<Insured> itr = insuredSet.iterator();
while (itr.hasNext()) {
itr.next();
itr.remove();
}
Moving to the next is important here as it should take the index to remove element.
List<String> list1 = new ArrayList<>();
list1.addAll(OriginalList);
List<String> list2 = new ArrayList<>();
list2.addAll(OriginalList);
This is also an option.
If your goal is to remove all elements from the list, you can iterate over each item, and then call:
list.clear()
What about of
import java.util.Collections;
List<A> abc = Collections.synchronizedList(new ArrayList<>());
ERROR
There was a mistake when I added to the same list from where I took elements:
fun <T> MutableList<T>.mathList(_fun: (T) -> T): MutableList<T> {
for (i in this) {
this.add(_fun(i)) <--- ERROR
}
return this <--- ERROR
}
DECISION
Works great when adding to a new list:
fun <T> MutableList<T>.mathList(_fun: (T) -> T): MutableList<T> {
val newList = mutableListOf<T>() <--- DECISION
for (i in this) {
newList.add(_fun(i)) <--- DECISION
}
return newList <--- DECISION
}
Just add a break after your ArrayList.remove(A) statement
I want to have a reversed list view on a list (in a similar way than List#sublist provides a sublist view on a list). Is there some function which provides this functionality?
I don't want to make any sort of copy of the list nor modify the list.
It would be enough if I could get at least a reverse iterator on a list in this case though.
Also, I know how to implement this myself. I'm just asking if Java already provides something like this.
Demo implementation:
static <T> Iterable<T> iterableReverseList(final List<T> l) {
return new Iterable<T>() {
public Iterator<T> iterator() {
return new Iterator<T>() {
ListIterator<T> listIter = l.listIterator(l.size());
public boolean hasNext() { return listIter.hasPrevious(); }
public T next() { return listIter.previous(); }
public void remove() { listIter.remove(); }
};
}
};
}
I just have found out that some List implementations have descendingIterator() which is what I need. Though there is no general such implementation for List. Which is kind of strange because the implementation I have seen in LinkedList is general enough to work with any List.
Use the .clone() method on your List. It will return a shallow copy, meaning that it will contain pointers to the same objects, so you won't have to copy the list. Then just use Collections.
Ergo,
Collections.reverse(list.clone());
If you are using a List and don't have access to clone() you can use subList():
List<?> shallowCopy = list.subList(0, list.size());
Collections.reverse(shallowCopy);
Guava provides this: Lists.reverse(List)
List<String> letters = ImmutableList.of("a", "b", "c");
List<String> reverseView = Lists.reverse(letters);
System.out.println(reverseView); // [c, b, a]
Unlike Collections.reverse, this is purely a view... it doesn't alter the ordering of elements in the original list. Additionally, with an original list that is modifiable, changes to both the original list and the view are reflected in the other.
If i have understood correct then it is one line of code .It worked for me .
Collections.reverse(yourList);
Its not exactly elegant, but if you use List.listIterator(int index) you can get a bi-directional ListIterator to the end of the list:
//Assume List<String> foo;
ListIterator li = foo.listIterator(foo.size());
while (li.hasPrevious()) {
String curr = li.previous();
}
I use this:
public class ReversedView<E> extends AbstractList<E>{
public static <E> List<E> of(List<E> list) {
return new ReversedView<>(list);
}
private final List<E> backingList;
private ReversedView(List<E> backingList){
this.backingList = backingList;
}
#Override
public E get(int i) {
return backingList.get(backingList.size()-i-1);
}
#Override
public int size() {
return backingList.size();
}
}
like this:
ReversedView.of(backingList) // is a fully-fledged generic (but read-only) list
java.util.Deque has descendingIterator() - if your List is a Deque, you can use that.
Collections.reverse(nums) ... It actually reverse the order of the elements.
Below code should be much appreciated -
List<Integer> nums = new ArrayList<Integer>();
nums.add(61);
nums.add(42);
nums.add(83);
nums.add(94);
nums.add(15);
//Tosort the collections uncomment the below line
//Collections.sort(nums);
Collections.reverse(nums);
System.out.println(nums);
Output: 15,94,83,42,61
I know this is an old post but today I was looking for something like this. In the end I wrote the code myself:
private List reverseList(List myList) {
List invertedList = new ArrayList();
for (int i = myList.size() - 1; i >= 0; i--) {
invertedList.add(myList.get(i));
}
return invertedList;
}
Not recommended for long Lists, this is not optimized at all. It's kind of an easy solution for controlled scenarios (the Lists I handle have no more than 100 elements).
Hope it helps somebody.
You can also invert the position when you request an object:
Object obj = list.get(list.size() - 1 - position);
For small sized list we can create LinkedList and then can make use of descending iterator as:
List<String> stringList = new ArrayList<>(Arrays.asList("One", "Two", "Three"));
stringList.stream().collect(Collectors.toCollection(LinkedList::new))
.descendingIterator().
forEachRemaining(System.out::println); // Three, Two, One
System.out.println(stringList); // One, Two, Three
You can also do this:
static ArrayList<String> reverseReturn(ArrayList<String> alist)
{
if(alist==null || alist.isEmpty())
{
return null;
}
ArrayList<String> rlist = new ArrayList<>(alist);
Collections.reverse(rlist);
return rlist;
}