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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;
}
I have a list like this
List contains set of dtime,uptime values.I want to get the list items i.e., dtime into one and
uptime into another variable.Likewise I want to get all the dtime and uptime pair values seperatly into
the variables using for loop in java.How can I achieve this.Is it possible list or vector?Please help me.
Pseudo code
List.get(0).get(0)-->gives 1st dtime
List.get(0).get(1)-->gives 1st uptime
List.get(1).get(0)-->gives 2nd dtime
List.get(1).get(1)-->gives 2nd uptime
And so on..
How to implement this with for loop I am not getting.I am new to java>please help me..
First Convert That ArrayList into Object[] array then get the value like given below code...driver_ModelsObj is an array convert that into drives object array then get the value from inside the array.
for(int indx=0;indx<driver_ModelsObj.size();indx++){
Object[] drivers=(Object[]) driver_ModelsObj.get(indx);
String Device_ID=drivers[0].toString();
}
If your list is as below
List list = [[1],[2],[3]];
We can retrieve the each value as below.
((List)list.get(0)).get(0); //This will retrieve value 1
((List)list.get(1)).get(0); //This will retrieve value 2
Sounds like you could use a domain object containing uptime and downtime.
For example,
public class Stats {
int dtime;
int uptime;
}
Then you can have a List<Stats> and access it like this:
mylist.get(0).dtime
mylist.get(0).uptime
mylist.get(1).dtime
mylist.get(1).uptime
Part of the (newer) Collcetions framework, List is almost always a better alternative than Vector
List.get(0).get(0)-->gives 1st dtime
List.get(0).get(1)-->gives 1st uptime
Well, what you're doing here, is getting the list at position 0, and getting item 1 from that list. In a for loop we can express this as:
for(int x = 0; x < List.size(); x++)
{
for(int y = 0; y < List.get(x).size(); y++)
{
if(y % 2 == 0)
{
// It's a dtime object.
}
else
{
// It's an uptime object.
}
}
}
Before this, you could declare some lists of your own:
List<DTime> listD = new ArrayList<ATimeObject>();
List<UpTime> listUp = new ArrayList<UpTime>();
Now when you're cycling through, all you need to do is add the relevant object:
if(y % 2 == 0)
{
listD.add(List.get(x).get(y));
}
else
{
listUp.add(List.get(x).get(y));
}
You should create a POJO like
public class TimeData {
double dtime;
Date uptime;
}
Then add each POJO to array list and then iterate it.
List<TimeData> oList = new ArrayList<TimeData>();
int nSize = oList.size();
for(int i=0;i<nSize;i++){
TimeData child = oList.get(i);
// get value using getters
}
You can try this ,Let say you have variables like
double dtime;
Timestamp tp;
And listofresults is coming from query results.
listofresults = results.getResultList();
If list is coming from query then put it in the loop this way in the condition of for loop
for(int i=0;i< listofresults.size() ;i=i+2)
{
dtime=(double) listofresults.get(i);
//cast because the value is of object type
tp=(TimeStamp) listofresults.get(i+1);
//cast because the value is of object type
//do something with these variables
}
I recommend creating a wrapper for it.
public class UpTimeDownTime {
MyTimeDataClass downtime;
MyTimeDataClass uptime;
public UpTimeDownTime(MyTimeDataClass downtime, MyTimeDataClass uptime){
this.downtime = downtime;
this.uptime = uptime;
}
public MyTimeDataClass getDowntime () {
return downtime;
}
public MyTimeDataClass getUptime () {
return uptime;
}
public static void main (String[] args) {
List<List<MyTimeDataClass>> List = ...;
List<UpTimeDownTime> uptimeDowntime = new ArrayList<UpTimeDownTime>();
for(List<MyTimeDataClass> timeList : List){
UpTimeDownTime u = new UpTimeDownTime(timeList.get(0), timeList.get(1));
uptimeDowntime.add(u);
}
}
}
considering there is a structure as described in http://prohost.lt/stack/q.png
In the main arrayList we have n elements(sub arrayLists) (as for image we can consider n=3).
How to make a dynamic loop thru all sub arrayLists so as to get:
sarraylist1.get(0), sarrayList2.get(0), .. sarraylistN.get(0)
In case n=3 it would be
for(sarraylist1){
for( sarraylist2){
for(sarraylist3){
element from every list
}
}
}
Maybe I misunderstood what you were asking, but would something as simple as this not work?
List<List> lists = new ArrayList<List>();
for(List l : lists){
for(List subl: l)
Object o = subl.get(0);
}
Based on your scheme of the image, you can also do it with the "for while loop":
for (int mainArrayElements = 0; mainArrayElements < listOfLists.size(); mainArrayElements++) {
List arraysInMain = ((List)listOfLists.get(mainArrayElements));
for (int subArrayElements = 0; subArrayElements < arraysInMain.size(); subArrayElements++) {
List deepArray = ((List)listOfLists.get(subArrayElements));
for (int innerArrayElements = 0; innerArrayElements < deepArray.size() ; innerArrayElements++) {
System.out.println(deepArray.get(innerArrayElements));
}
}
}
But as mentioned before, recursion would be a really great combo to knock out the problem :).
Best regards.
If I understood correctly, you have one List containing n Lists for which you want to get the first, second and so on, element of each list. This code does that.
for (int i = 0; i < highestSizeOfChildrenList; i++) {
List<Object> lstOfItemInChild = new Arraylist<Object>();
for (List childList : mainList) {
if (i < childList.size()) {
lstOfItemInChild.add(childList.get(i));
}
}
// lstOfItemInChild now contains all the element i of the sub arrays
}
You will need to know the size of the biggest subList to get all the elements of that list, that is what highestSizeOfChildrenList represents.
After watching your comment I suggest you to use recursive approach to solve your problem.
public void recursiveCall(ArrayList list)
{
for ( Object obj : list)
{
if (obj instanceOf ArrayList)
{
recursiveCall((ArrayList)obj);
}
else
{
System.out.print(obj);
}
}
}
I got five ArrayLists with different lengths but when I concatenate all the array lists, I get an array index out of bound exception as the items in a all the array list are not the same. I want to either include null when the item not found or make the array list of same size. How to achieve this in android.. please help me with some example and thanks in advance
arraylist 1 = 25 items
attaylist 2 = 35 items
arraylits 3 = 5 items
arraylist 4 = 13 items
arraylist 5 = 40 items.
when i display all items in one view (eg.web view) i could scroll upto 5 items when scroll to 6th item i get exception. how to overcome this. please help
here is my code whre i parse and add to lists.
I don't know but see if this helps:
package your.package
import java.util.ArrayList;
import android.app.Activity;
import android.os.Bundle;
import android.widget.TextView;
public class TestProjectActivity extends Activity {
#Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
TextView tv = new TextView(this);
setContentView(tv);
ArrayList<Integer> a = new ArrayList<Integer>();
a.add(1);
a.add(2);
a.add(3);
a.add(4);
ArrayList<Integer> b = new ArrayList<Integer>();
b.add(5);
b.add(6);
b.add(7);
a.addAll(b);
String p = "";
for (int i = 0; i < a.size(); i++)
{
p = String.format("%s, %d", p, a.get(i));
}
tv.setText(p);
}
}
It is easier to see what the problem is if your code is posted. If you want to return null when you get an out of bounds exception, though, something like this should do it:
try {
//your code that's throwing index out of bounds
//this block should include as little of your code as possible
//for example, don't put a for loop in here --
// put this inside the for loop if you can.
} catch (IndexOutOfBoundsException e) {
return null;
// or do what you need to do with the null
}
Here's some introductory material on exception handling in case you'd like to check it out.
this is not android.. its pure java issue
technically , arraylist is a kind of mutable array...
just check where the index you are calling is less than the size of arrayList if yes,go concatenating, if not so its you are out bound exception..
or you can try to convert your arraylist to array like this (suppose data type is String):
String mArray[]=mArrayList.toArray(new String[SIZE]);
this will assign null to empty elements in case mArrayList size < mArray
good luck
Make some Default values in your class
like
public class BOOK {
String BId="";
String BTitle="";
String prelimTitle="";
String prelimIntro="";
String prelimInsight="";
}
I would recommend to have another data structure, and add list of that item into main Book Structure, new Datastructre may be:
case XmlPullParser.END_TAG:
if (xmlpullparser.getName().toString().equals("BOOK")) {
book.setInsightList(insightList);
book.setcasestudyList(casestudyList);
book.setquotesList(quoteList);
book.setoneminwonderList(oneminwonderList);
book.setsecretList(secretList);
bookList.add(book);
Intro.book = book;
} else if (xmlpullparser.getName().toString().equals("chapter")) {
if(objInsight==null)
objInsight=new INSIGHT();
listInsights.add(objInsight);
objInsight=null;
} else if (xmlpullparser.getName().toString().equals("secret")) {
secretList.add(secret);
} else if (xmlpullparser.getName().toString().equals("prelim")) {
secretList.add(secret);
this.prelim = false;
} else if (xmlpullparser.getName().toString().equals("prelimTitle")) {
this.prelimTitle = false;
} else if (xmlpullparser.getName().toString().equals("prelimIntro")) {
this.prelimIntro = false;
} else if (xmlpullparser.getName().toString()
.equals("prelimInsight")) {
this.prelimInsight = false;
} else if (xmlpullparser.getName().toString()
.equals("prelimContent")) {
this.prelimContent = false;
} else if (xmlpullparser.getName().toString()
.equals("chapterIntro")) {
secretList.add(secret);
this.chapterIntro = false;
} else if (xmlpullparser.getName().toString().equals("bold")) {
this.chapterIntrobold = false;
} else if (xmlpullparser.getName().toString().equals("secretIntro")) {
this.secretIntro = false;
} else if (xmlpullparser.getName().toString().equals("insight")) {
} else if (xmlpullparser.getName().toString().equals("caseStudy")) {
this.caseStudy = false;
casestudyList.add(casestudy);
} else if (xmlpullparser.getName().toString().equals("oneMinute")) {
this.oneMinute = false;
oneminwonderList.add(oneminwonder);
} else if (xmlpullparser.getName().toString().equals("quote")) {
quoteList.add(quotes);
} else if (xmlpullparser.getName().toString()
.equals("quoteContent")) {
this.quoteContent = false;
} else if (xmlpullparser.getName().toString()
.equals("quoteCitation")) {
this.quoteCitation = false;
} else if (xmlpullparser.getName().toString()
.equals("secretContent")) {
this.secretContent = false;
}
break;
You got that exception there may be not enough space in to thst arraylist.
Supose there are three arraylist like a with size 10,b with size 10 and c with size 10.
Now you are adding the that all arraylist to the ArrayList z with size 10 then there are total 30 data but the index are only 10. So it cause the IndexOutofBount exception.
Solution
So to solve that you have to take the new ArrayList z with the size like sum of that all three arraylist size.
and then add the three arrayList data to that arrayList.
You can give the Size to the ArrayList with below code.
ArrayList<Integer>[size] arrayOfLists = new ArrayList<Integer>[size]();
Hope it will help you.
If not then let me know.
Enjoy. :))
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;
}