Develop Reference

How to check if two objects in a ArrayList are the same? [duplicate]

How could I go about detecting (returning true/false) whether an ArrayList contains more than one of the same element in Java?
Many thanks,
Terry
Edit
Forgot to mention that I am not looking to compare "Blocks" with each other but their integer values. Each "block" has an int and this is what makes them different.
I find the int of a particular Block by calling a method named "getNum" (e.g. table1[0][2].getNum();

Simplest: dump the whole collection into a Set (using the Set(Collection) constructor or Set.addAll), then see if the Set has the same size as the ArrayList.
List<Integer> list = ...;
Set<Integer> set = new HashSet<Integer>(list);
if(set.size() < list.size()){
/* There are duplicates */
}
Update: If I'm understanding your question correctly, you have a 2d array of Block, as in
Block table[][];
and you want to detect if any row of them has duplicates?
In that case, I could do the following, assuming that Block implements "equals" and "hashCode" correctly:
for (Block[] row : table) {
Set set = new HashSet<Block>();
for (Block cell : row) {
set.add(cell);
}
if (set.size() < 6) { //has duplicate
}
}
I'm not 100% sure of that for syntax, so it might be safer to write it as
for (int i = 0; i < 6; i++) {
Set set = new HashSet<Block>();
for (int j = 0; j < 6; j++)
set.add(table[i][j]);
...
Set.add returns a boolean false if the item being added is already in the set, so you could even short circuit and bale out on any add that returns false if all you want to know is whether there are any duplicates.

Improved code, using return value of Set#add instead of comparing the size of list and set.
public static <T> boolean hasDuplicate(Iterable<T> all) {
Set<T> set = new HashSet<T>();
// Set#add returns false if the set does not change, which
// indicates that a duplicate element has been added.
for (T each: all) if (!set.add(each)) return true;
return false;
}

With Java 8+ you can use Stream API:
boolean areAllDistinct(List<Block> blocksList) {
return blocksList.stream().map(Block::getNum).distinct().count() == blockList.size();
}

If you are looking to avoid having duplicates at all, then you should just cut out the middle process of detecting duplicates and use a Set.

Improved code to return the duplicate elements
Can find duplicates in a Collection
return the set of duplicates
Unique Elements can be obtained from the Set
public static <T> List getDuplicate(Collection<T> list) {
final List<T> duplicatedObjects = new ArrayList<T>();
Set<T> set = new HashSet<T>() {
#Override
public boolean add(T e) {
if (contains(e)) {
duplicatedObjects.add(e);
}
return super.add(e);
}
};
for (T t : list) {
set.add(t);
}
return duplicatedObjects;
}
public static <T> boolean hasDuplicate(Collection<T> list) {
if (getDuplicate(list).isEmpty())
return false;
return true;
}

I needed to do a similar operation for a Stream, but couldn't find a good example. Here's what I came up with.
public static <T> boolean areUnique(final Stream<T> stream) {
final Set<T> seen = new HashSet<>();
return stream.allMatch(seen::add);
}
This has the advantage of short-circuiting when duplicates are found early rather than having to process the whole stream and isn't much more complicated than just putting everything in a Set and checking the size. So this case would roughly be:
List<T> list = ...
boolean allDistinct = areUnique(list.stream());

If your elements are somehow Comparable (the fact that the order has any real meaning is indifferent -- it just needs to be consistent with your definition of equality), the fastest duplicate removal solution is going to sort the list ( 0(n log(n)) ) then to do a single pass and look for repeated elements (that is, equal elements that follow each other) (this is O(n)).
The overall complexity is going to be O(n log(n)), which is roughly the same as what you would get with a Set (n times long(n)), but with a much smaller constant. This is because the constant in sort/dedup results from the cost of comparing elements, whereas the cost from the set is most likely to result from a hash computation, plus one (possibly several) hash comparisons. If you are using a hash-based Set implementation, that is, because a Tree based is going to give you a O( n log²(n) ), which is even worse.
As I understand it, however, you do not need to remove duplicates, but merely test for their existence. So you should hand-code a merge or heap sort algorithm on your array, that simply exits returning true (i.e. "there is a dup") if your comparator returns 0, and otherwise completes the sort, and traverse the sorted array testing for repeats. In a merge or heap sort, indeed, when the sort is completed, you will have compared every duplicate pair unless both elements were already in their final positions (which is unlikely). Thus, a tweaked sort algorithm should yield a huge performance improvement (I would have to prove that, but I guess the tweaked algorithm should be in the O(log(n)) on uniformly random data)

If you want the set of duplicate values:
import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
public class FindDuplicateInArrayList {
public static void main(String[] args) {
Set<String> uniqueSet = new HashSet<String>();
List<String> dupesList = new ArrayList<String>();
for (String a : args) {
if (uniqueSet.contains(a))
dupesList.add(a);
else
uniqueSet.add(a);
}
System.out.println(uniqueSet.size() + " distinct words: " + uniqueSet);
System.out.println(dupesList.size() + " dupesList words: " + dupesList);
}
}
And probably also think about trimming values or using lowercase ... depending on your case.

Simply put:
1) make sure all items are comparable
2) sort the array
2) iterate over the array and find duplicates

To know the Duplicates in a List use the following code:It will give you the set which contains duplicates.
public Set<?> findDuplicatesInList(List<?> beanList) {
System.out.println("findDuplicatesInList::"+beanList);
Set<Object> duplicateRowSet=null;
duplicateRowSet=new LinkedHashSet<Object>();
for(int i=0;i<beanList.size();i++){
Object superString=beanList.get(i);
System.out.println("findDuplicatesInList::superString::"+superString);
for(int j=0;j<beanList.size();j++){
if(i!=j){
Object subString=beanList.get(j);
System.out.println("findDuplicatesInList::subString::"+subString);
if(superString.equals(subString)){
duplicateRowSet.add(beanList.get(j));
}
}
}
}
System.out.println("findDuplicatesInList::duplicationSet::"+duplicateRowSet);
return duplicateRowSet;
}

best way to handle this issue is to use a HashSet :
ArrayList<String> listGroupCode = new ArrayList<>();
listGroupCode.add("A");
listGroupCode.add("A");
listGroupCode.add("B");
listGroupCode.add("C");
HashSet<String> set = new HashSet<>(listGroupCode);
ArrayList<String> result = new ArrayList<>(set);
Just print result arraylist and see the result without duplicates :)

This answer is wrriten in Kotlin, but can easily be translated to Java.
If your arraylist's size is within a fixed small range, then this is a great solution.
var duplicateDetected = false
if(arrList.size > 1){
for(i in 0 until arrList.size){
for(j in 0 until arrList.size){
if(i != j && arrList.get(i) == arrList.get(j)){
duplicateDetected = true
}
}
}
}

private boolean isDuplicate() {
for (int i = 0; i < arrayList.size(); i++) {
for (int j = i + 1; j < arrayList.size(); j++) {
if (arrayList.get(i).getName().trim().equalsIgnoreCase(arrayList.get(j).getName().trim())) {
return true;
}
}
}
return false;
}

String tempVal = null;
for (int i = 0; i < l.size(); i++) {
tempVal = l.get(i); //take the ith object out of list
while (l.contains(tempVal)) {
l.remove(tempVal); //remove all matching entries
}
l.add(tempVal); //at last add one entry
}
Note: this will have major performance hit though as items are removed from start of the list.
To address this, we have two options. 1) iterate in reverse order and remove elements. 2) Use LinkedList instead of ArrayList. Due to biased questions asked in interviews to remove duplicates from List without using any other collection, above example is the answer. In real world though, if I have to achieve this, I will put elements from List to Set, simple!

/**
* Method to detect presence of duplicates in a generic list.
* Depends on the equals method of the concrete type. make sure to override it as required.
*/
public static <T> boolean hasDuplicates(List<T> list){
int count = list.size();
T t1,t2;
for(int i=0;i<count;i++){
t1 = list.get(i);
for(int j=i+1;j<count;j++){
t2 = list.get(j);
if(t2.equals(t1)){
return true;
}
}
}
return false;
}
An example of a concrete class that has overridden equals() :
public class Reminder{
private long id;
private int hour;
private int minute;
public Reminder(long id, int hour, int minute){
this.id = id;
this.hour = hour;
this.minute = minute;
}
#Override
public boolean equals(Object other){
if(other == null) return false;
if(this.getClass() != other.getClass()) return false;
Reminder otherReminder = (Reminder) other;
if(this.hour != otherReminder.hour) return false;
if(this.minute != otherReminder.minute) return false;
return true;
}
}

ArrayList<String> withDuplicates = new ArrayList<>();
withDuplicates.add("1");
withDuplicates.add("2");
withDuplicates.add("1");
withDuplicates.add("3");
HashSet<String> set = new HashSet<>(withDuplicates);
ArrayList<String> withoutDupicates = new ArrayList<>(set);
ArrayList<String> duplicates = new ArrayList<String>();
Iterator<String> dupIter = withDuplicates.iterator();
while(dupIter.hasNext())
{
String dupWord = dupIter.next();
if(withDuplicates.contains(dupWord))
{
duplicates.add(dupWord);
}else{
withoutDupicates.add(dupWord);
}
}
System.out.println(duplicates);
System.out.println(withoutDupicates);

A simple solution for learners.
//Method to find the duplicates.
public static List<Integer> findDublicate(List<Integer> numList){
List<Integer> dupLst = new ArrayList<Integer>();
//Compare one number against all the other number except the self.
for(int i =0;i<numList.size();i++) {
for(int j=0 ; j<numList.size();j++) {
if(i!=j && numList.get(i)==numList.get(j)) {
boolean isNumExist = false;
//The below for loop is used for avoid the duplicate again in the result list
for(Integer aNum: dupLst) {
if(aNum==numList.get(i)) {
isNumExist = true;
break;
}
}
if(!isNumExist) {
dupLst.add(numList.get(i));
}
}
}
}
return dupLst;
}

compare Object attribute value inside list

Class Order
{
String name;
Order(String n)
{ name = n; }
//setter and getters of name
}
Order a = new Order("same");
Order b = new Order("same");
Order c = new Order("diff");
List<Order> nameList// a,b,c
I want to
seperate list of Orders
List<Order> dupList// a,b
List<Order> nondupList// c
Now I want to check whether same name is available in multiple orders of "nameList".
I achieved that using index of List and compare with other than that index List Orders.
But is there any other better way to achieve this.

Probably one other way could be - Override hashCode method and equals method. Generate hasCode on calculation of string name.
public class Order {
String name;
public Order(String n) {
name = n;
}
// setter and getters of name
#Override
public int hashCode() {
int h = 0;
int len = name.length();
for (int i = 0; i < len; i++)
h = 31 * h + name.charAt(i);
return h;
}
#Override
public boolean equals(Object obj) {
if(obj == null)
return false;
else if(this.hashCode() == obj.hashCode())
return true;
return false;
}
}
...
List<Order> nameList = ...;// a,b,c
Set<Order> nonDuplicate= new HashSet<Order>(nameList);

If you want to use pure java, add the elements to the a List, and sort it with the appropriate comparator. Then iterate over the list, keeping track of the previous element, doing a control break; in other words, if the element is the same as the previous both are a duplicate. If they are not (or it is the first), they are a candidate and you need to wait for the next check to find a duplicate.
If you don't want to sort, you can add the elements to a Set as they appear; if before adding an element it is already in the set, you can add it to the duplicate set. You can do the check on both sets removing as you go, or remove from the complete set the duplicates at the end. You can use any collection, but Set is more efficient since it has a fast contains method.
If you can use libraries, you can just use Guava and add everything to a multiset (http://google-collections.googlecode.com/svn/trunk/javadoc/com/google/common/collect/Multiset.html ) Then iterate over the multiset and you have the count per element.

You could use Map>, get list for given name, if null, create it and put it in, add current order in that list.

Avoiding printing of duplicates in ArrayList

I have one array list contains object( id,name).need to remove if duplicate names in the list
I need to print only name once if it is repeating

Use SET if you want don't want duplicate entries in your list:
HashSet if you don't required sequence
LinkedHashSet if you need to print name in sequence
Note: You must override equals and hashcode in object(ID,Name) otherwise you will find duplicate object in SET

public static boolean checkDuplicate(ArrayList list) {
HashSet set = new HashSet();
for (int i = 0; i < list.size(); i++) {
boolean val = set.add(list.get(i));
if (val == false) {
return val;
}
}
return true;
}

Use a java.util.Set instead. It cannot contain duplicates.
Reference: http://docs.oracle.com/javase/6/docs/api/java/util/Set.html
Set<String> myNames = new HashSet<String>();
myNames.add("JAMES");
myNames.add("ELLA");
myNames.add("JAMES");
myNames.size() // will return 2

Use a Set data structure with implementing appropriate Comparator. Run a loop and put your objects in that set. Print this set.

Arraylist provides an iterator that allows to delete the last element returned by the iterator from the list itself. So we just keep track of the names that we've already seen. A Set implementation is helpful.
Set<String> names = new HashSet<>();
Iterator<MyType> it = getListFromSomewhere().iterator(); // you know how
while(it.hasNext()) {
if (!names.add(it.next().getName()) {
it.remove();
}
}
Hint: adding to a set returns false, if the value is already in the set.
Remarks
Your requirement wasn't that clear: here we keep the first entry with a name and delete all following entries from the list. If the list is not sorted by id, then it is rather random, which entries are kept an which are removed.

You can do it with the help of Set as well as List family
We have seen how to remove duplicates with Set family. We can do it with the help of List family as well. Please refer below code.
package com.rais.util;
import java.util.ArrayList;
import java.util.List;
/**
* #author Rais.Alam
* #project Utils
* #date Dec 18, 2012
*/
public class ConverterUtil
{
/**
* #param args
*/
public static void main(String[] args)
{
List<String> original = new ArrayList<String>();
original.add("Tomcat");
original.add("Jboss");
original.add("GlassFish");
original.add("Weblogic");
original.add("Tomcat");
List<String> newList = removeDuplicates(original);
for (String value : newList)
{
System.out.println(value);
}
}
public static List<String> removeDuplicates(List<String> original)
{
List<String> tempList = new ArrayList<String>();
for (String value : original)
{
if (!tempList.contains(value))
{
tempList.add(value);
}
}
return tempList;
}
}

how to get duplicated and non duplicated element of arrayList?

I have an object as Riziv with three variables as id, cnk and product. Then I search in a databank for this object and add it to a ArrayList as ArrayList<Riziv> list.
Now I should checkout if all object in his array are the same cnk then return true otherwise I should return all objects which are not the same cnk with error message.
public class Riziv{ String id, cnk, product; }
ArrayList<Riziv> list = getArrayListFromDatabank(id);
public void getDuplicatedWhichHasTheSameCnk(){
}
}

Using standard JVM structures (MultiMap is provided by guava), you can do that:
public List<Riviz> getDuplicates(final List<Riviz> l)
{
final HashMap<String, List<Riviz>> m = new HashMap<String, List<Riviz>>();
final List<Riviz> ret = new ArrayList<Riviz>();
String cnk;
for (final Riviz r: l) {
cnk = r.getCnk();
if (!m.contains(cnk))
m.add(cnk, new ArrayList<Riviz>());
m.get(cnk).add(r);
}
List<Riviz> tmp;
for (final Map.Entry<String, List<Riviz>> entry: m.entrySet()) {
tmp = entry.getValue();
if (tmp.size() == 1) // no dups
continue;
ret.addAll(tmp);
}
return ret;
}
ret will contain the duplicates. You can change that function to return a Map<String, Riviz> instead, and filter out entries where the list size is only one. You'll then get a map with the conflicting cnks as keys and a list of dups as values.

I am not clear exactly what you want however I suspect you want something like this.
MultiMap<Key, Riziv> multiMap =
List<Riziv> list =
for(Riziv r: list)
multiMap.put(r.getCnk(), r);
for(Key cnk: multiMap.keySet()) {
Collection<Riziv> sameCnk = multiMap.get(cnk);
// check size and compare entries
}
The multi-map will have the list of Riziv objects for each Cnk.

One way to do it is write a comparator to sort the list by cnk String and then compare each consecutive cnk String to the next, if you find a duplicate, they will be right next to eachother.
1.) Sort the list using a comparator by sorting on the cnk variable.
2.) Compare each element in the list to the next for duplicates.
There's probably many other ways to solve this, this is just the first that came to mind.
I did not test this so you have been forewarned lol:
ArrayList<Riziv> rizArray = new ArrayList<Riziv>();
//Sort the array by the CNK variable.
Collections.sort(rizArray, new Comparator<Riziv>(){
#Override
public int compare(Riziv arg0, Riziv arg1) {
//Return the comparison of the Strings.
//Use .compareToIgnoreCase if you want to ignore upper/lower case.
return arg0.getCnk().compareTo(arg1.getCnk());
}
});
//List should be in alphabetical order at this point.
List<Riziv> duplicates = new ArrayList<Riziv>();
Riziv rizPrevious = null;
for(Riziv riz: rizArray){
if(rizPrevious == null){
rizPrevious = riz;
continue;
}
if(riz.getCnk().compareTo(rizPrevious.getCnk()) == 0){
duplicates.add(riz);
}
rizPrevious = riz;
}

Java: Detect duplicates in ArrayList?

How could I go about detecting (returning true/false) whether an ArrayList contains more than one of the same element in Java?
Many thanks,
Terry
Edit
Forgot to mention that I am not looking to compare "Blocks" with each other but their integer values. Each "block" has an int and this is what makes them different.
I find the int of a particular Block by calling a method named "getNum" (e.g. table1[0][2].getNum();

Simplest: dump the whole collection into a Set (using the Set(Collection) constructor or Set.addAll), then see if the Set has the same size as the ArrayList.
List<Integer> list = ...;
Set<Integer> set = new HashSet<Integer>(list);
if(set.size() < list.size()){
/* There are duplicates */
}
Update: If I'm understanding your question correctly, you have a 2d array of Block, as in
Block table[][];
and you want to detect if any row of them has duplicates?
In that case, I could do the following, assuming that Block implements "equals" and "hashCode" correctly:
for (Block[] row : table) {
Set set = new HashSet<Block>();
for (Block cell : row) {
set.add(cell);
}
if (set.size() < 6) { //has duplicate
}
}
I'm not 100% sure of that for syntax, so it might be safer to write it as
for (int i = 0; i < 6; i++) {
Set set = new HashSet<Block>();
for (int j = 0; j < 6; j++)
set.add(table[i][j]);
...
Set.add returns a boolean false if the item being added is already in the set, so you could even short circuit and bale out on any add that returns false if all you want to know is whether there are any duplicates.

Improved code, using return value of Set#add instead of comparing the size of list and set.
public static <T> boolean hasDuplicate(Iterable<T> all) {
Set<T> set = new HashSet<T>();
// Set#add returns false if the set does not change, which
// indicates that a duplicate element has been added.
for (T each: all) if (!set.add(each)) return true;
return false;
}

With Java 8+ you can use Stream API:
boolean areAllDistinct(List<Block> blocksList) {
return blocksList.stream().map(Block::getNum).distinct().count() == blockList.size();
}

If you are looking to avoid having duplicates at all, then you should just cut out the middle process of detecting duplicates and use a Set.

Improved code to return the duplicate elements
Can find duplicates in a Collection
return the set of duplicates
Unique Elements can be obtained from the Set
public static <T> List getDuplicate(Collection<T> list) {
final List<T> duplicatedObjects = new ArrayList<T>();
Set<T> set = new HashSet<T>() {
#Override
public boolean add(T e) {
if (contains(e)) {
duplicatedObjects.add(e);
}
return super.add(e);
}
};
for (T t : list) {
set.add(t);
}
return duplicatedObjects;
}
public static <T> boolean hasDuplicate(Collection<T> list) {
if (getDuplicate(list).isEmpty())
return false;
return true;
}

I needed to do a similar operation for a Stream, but couldn't find a good example. Here's what I came up with.
public static <T> boolean areUnique(final Stream<T> stream) {
final Set<T> seen = new HashSet<>();
return stream.allMatch(seen::add);
}
This has the advantage of short-circuiting when duplicates are found early rather than having to process the whole stream and isn't much more complicated than just putting everything in a Set and checking the size. So this case would roughly be:
List<T> list = ...
boolean allDistinct = areUnique(list.stream());

If your elements are somehow Comparable (the fact that the order has any real meaning is indifferent -- it just needs to be consistent with your definition of equality), the fastest duplicate removal solution is going to sort the list ( 0(n log(n)) ) then to do a single pass and look for repeated elements (that is, equal elements that follow each other) (this is O(n)).
The overall complexity is going to be O(n log(n)), which is roughly the same as what you would get with a Set (n times long(n)), but with a much smaller constant. This is because the constant in sort/dedup results from the cost of comparing elements, whereas the cost from the set is most likely to result from a hash computation, plus one (possibly several) hash comparisons. If you are using a hash-based Set implementation, that is, because a Tree based is going to give you a O( n log²(n) ), which is even worse.
As I understand it, however, you do not need to remove duplicates, but merely test for their existence. So you should hand-code a merge or heap sort algorithm on your array, that simply exits returning true (i.e. "there is a dup") if your comparator returns 0, and otherwise completes the sort, and traverse the sorted array testing for repeats. In a merge or heap sort, indeed, when the sort is completed, you will have compared every duplicate pair unless both elements were already in their final positions (which is unlikely). Thus, a tweaked sort algorithm should yield a huge performance improvement (I would have to prove that, but I guess the tweaked algorithm should be in the O(log(n)) on uniformly random data)

If you want the set of duplicate values:
import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
public class FindDuplicateInArrayList {
public static void main(String[] args) {
Set<String> uniqueSet = new HashSet<String>();
List<String> dupesList = new ArrayList<String>();
for (String a : args) {
if (uniqueSet.contains(a))
dupesList.add(a);
else
uniqueSet.add(a);
}
System.out.println(uniqueSet.size() + " distinct words: " + uniqueSet);
System.out.println(dupesList.size() + " dupesList words: " + dupesList);
}
}
And probably also think about trimming values or using lowercase ... depending on your case.

Simply put:
1) make sure all items are comparable
2) sort the array
2) iterate over the array and find duplicates

To know the Duplicates in a List use the following code:It will give you the set which contains duplicates.
public Set<?> findDuplicatesInList(List<?> beanList) {
System.out.println("findDuplicatesInList::"+beanList);
Set<Object> duplicateRowSet=null;
duplicateRowSet=new LinkedHashSet<Object>();
for(int i=0;i<beanList.size();i++){
Object superString=beanList.get(i);
System.out.println("findDuplicatesInList::superString::"+superString);
for(int j=0;j<beanList.size();j++){
if(i!=j){
Object subString=beanList.get(j);
System.out.println("findDuplicatesInList::subString::"+subString);
if(superString.equals(subString)){
duplicateRowSet.add(beanList.get(j));
}
}
}
}
System.out.println("findDuplicatesInList::duplicationSet::"+duplicateRowSet);
return duplicateRowSet;
}

best way to handle this issue is to use a HashSet :
ArrayList<String> listGroupCode = new ArrayList<>();
listGroupCode.add("A");
listGroupCode.add("A");
listGroupCode.add("B");
listGroupCode.add("C");
HashSet<String> set = new HashSet<>(listGroupCode);
ArrayList<String> result = new ArrayList<>(set);
Just print result arraylist and see the result without duplicates :)

This answer is wrriten in Kotlin, but can easily be translated to Java.
If your arraylist's size is within a fixed small range, then this is a great solution.
var duplicateDetected = false
if(arrList.size > 1){
for(i in 0 until arrList.size){
for(j in 0 until arrList.size){
if(i != j && arrList.get(i) == arrList.get(j)){
duplicateDetected = true
}
}
}
}

private boolean isDuplicate() {
for (int i = 0; i < arrayList.size(); i++) {
for (int j = i + 1; j < arrayList.size(); j++) {
if (arrayList.get(i).getName().trim().equalsIgnoreCase(arrayList.get(j).getName().trim())) {
return true;
}
}
}
return false;
}

String tempVal = null;
for (int i = 0; i < l.size(); i++) {
tempVal = l.get(i); //take the ith object out of list
while (l.contains(tempVal)) {
l.remove(tempVal); //remove all matching entries
}
l.add(tempVal); //at last add one entry
}
Note: this will have major performance hit though as items are removed from start of the list.
To address this, we have two options. 1) iterate in reverse order and remove elements. 2) Use LinkedList instead of ArrayList. Due to biased questions asked in interviews to remove duplicates from List without using any other collection, above example is the answer. In real world though, if I have to achieve this, I will put elements from List to Set, simple!

/**
* Method to detect presence of duplicates in a generic list.
* Depends on the equals method of the concrete type. make sure to override it as required.
*/
public static <T> boolean hasDuplicates(List<T> list){
int count = list.size();
T t1,t2;
for(int i=0;i<count;i++){
t1 = list.get(i);
for(int j=i+1;j<count;j++){
t2 = list.get(j);
if(t2.equals(t1)){
return true;
}
}
}
return false;
}
An example of a concrete class that has overridden equals() :
public class Reminder{
private long id;
private int hour;
private int minute;
public Reminder(long id, int hour, int minute){
this.id = id;
this.hour = hour;
this.minute = minute;
}
#Override
public boolean equals(Object other){
if(other == null) return false;
if(this.getClass() != other.getClass()) return false;
Reminder otherReminder = (Reminder) other;
if(this.hour != otherReminder.hour) return false;
if(this.minute != otherReminder.minute) return false;
return true;
}
}

ArrayList<String> withDuplicates = new ArrayList<>();
withDuplicates.add("1");
withDuplicates.add("2");
withDuplicates.add("1");
withDuplicates.add("3");
HashSet<String> set = new HashSet<>(withDuplicates);
ArrayList<String> withoutDupicates = new ArrayList<>(set);
ArrayList<String> duplicates = new ArrayList<String>();
Iterator<String> dupIter = withDuplicates.iterator();
while(dupIter.hasNext())
{
String dupWord = dupIter.next();
if(withDuplicates.contains(dupWord))
{
duplicates.add(dupWord);
}else{
withoutDupicates.add(dupWord);
}
}
System.out.println(duplicates);
System.out.println(withoutDupicates);

A simple solution for learners.
//Method to find the duplicates.
public static List<Integer> findDublicate(List<Integer> numList){
List<Integer> dupLst = new ArrayList<Integer>();
//Compare one number against all the other number except the self.
for(int i =0;i<numList.size();i++) {
for(int j=0 ; j<numList.size();j++) {
if(i!=j && numList.get(i)==numList.get(j)) {
boolean isNumExist = false;
//The below for loop is used for avoid the duplicate again in the result list
for(Integer aNum: dupLst) {
if(aNum==numList.get(i)) {
isNumExist = true;
break;
}
}
if(!isNumExist) {
dupLst.add(numList.get(i));
}
}
}
}
return dupLst;
}