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I want to duplicate the elements in a linked list. This is what I have tried:
public class duplicate {
public static void main(String[] args) {
LinkedList <Integer> list = new LinkedList<Integer>() ;
list.add(2);
list.add(3);
list.add(4);
list.add(1);
list.add(0);
for( int i= 0 ; i<list.size(); i++) {
list.addAll(list);
System.out.println(list);
break;
}
}
}
But I got an infinite loop.
Firstly, your code runs just fine (the break prevents the infinite loop, see JLS-The break Statement).
Now, you don't need to for-loop over the list because List.addAll already
Appends all of the elements in the specified collection to the end of
this list, in the order that they are returned by the specified
collection's iterator (optional operation) (...)
So, just by doing this you're fine:
LinkedList <Integer> list = new LinkedList<Integer>() ;
//... code omitted (adds every number)
list.addAll(list);
However, if you want to use List.add instead of List.addAll you can do it like this (need to use for-loop):
LinkedList <Integer> list = new LinkedList<Integer>() ;
//... code omitted (adds every number)
int initialSize = list.size();
for( int i = 0 ; i < initialSize; i++) {
list.add(list.get(i));
}
System.out.println(list);
you are adding the list elements again and again in the for loop.
for(int i= 0 ; i < list.size(); i++) {
list.addAll(list);
System.out.println(list);
}
Every time it will grow & which leads the size to grow for each iteration. correct the step. Either use a local variable to store the size or change your logic
You can simply do list.addAll(list);.
If you want to use the add method as an exercise, you need to be careful to save the original size of the list before you start iterating. You can do it in the initialization part of your for loop:
public static void main(String[] args) {
List<Integer> list = new ArrayList<>(Arrays.asList(2, 3, 4, 1, 0));
for (int i = 0, size = list.size(); i < size; i++)
list.add(list.get(i));
System.out.println(list);
assert list.equals(Arrays.asList(2, 3, 4, 1, 0, 2, 3, 4, 1, 0));
}
Now you'll notice the above uses ArrayList, rather than LinkedList. In general, you should prefer ArrayList. Even the author of Java's LinkedList says he doesn't use it. See also this SO question about ArrayList vs LinkedList.
If you have to use LinkedList, you can just replace the second line from the above to this:
List<Integer> list = new LinkedList<>(Arrays.asList(2, 3, 4, 1, 0));
The rest of the code can remain unchanged. However, if you have a very long list, then using the get(index) method of the LinkedList class can reduce performance. With LinkedList, you get higher performance if you iterate (using LinkedList.iterator() or an enhanced for loop, than by using a plain for loop with get() calls. But you can't iterate over the list while adding to it, as you'll then get a ConcurrentModificationException. Instead, you can copy the linked list to an array and iterate over that, while adding to the list:
public static void main(String[] args) {
List<Integer> list = new LinkedList<>(Arrays.asList(2, 3, 4, 1, 0));
for (Integer element : list.toArray(new Integer[0])) {
list.add(element);
}
System.out.println(list);
assert list.equals(Arrays.asList(2, 3, 4, 1, 0, 2, 3, 4, 1, 0));
}
public static int[] convertListToArray(List<Integer> listResult) {
int[] result = new int[listResult.size()];
int i= 0;
for (int num : listResult) {
result[i++] = num;
}
return result;
}
Is there an efficient way to convert List to array without iterating List explicitly ?
Maybe it is possible by using methods like:
Arrays.copyOf(int [] origin , int newLength );
System.arraycopy(Object src, int srcPos,
Object dest, int destPos,
int length);
I know that there is a solution described here. However, I particularly interested in an efficient way of converting List<Integer> to int[]
Given the need to convert from Integer to int, I don't think you're going to find something more efficient than what you have, if I assume you're talking about runtime efficiency.
You might find converting to Integer[] first and then looping might be more efficient (below), but you might not, too. You'd have to test it in your specific scenario and see.
Here's that example:
int size = listResult.size();
int[] result = new int[size];
Integer[] temp = listResult.toArray(new Integer[size]);
for (int n = 0; n < size; ++n) {
result[n] = temp[n];
}
If efficiency is your primary concern, I think you can use your solution and make it more efficient by using an indexed for loop on the listResult if it is RandomAccess. However this makes the code much less readable, and you'd have to benchmark it for your use cases to see if it is more efficient.
public static int[] convertListToArray(List<Integer> listResult) {
int size = listResult.size();
int[] result = new int[size];
if (listResult instanceof RandomAccess)
{
for (int i = 0; i < size; i++)
{
result[i] = listResult.get(i);
}
}
else
{
int i = 0;
for (int num : listResult) {
result[i++] = num;
}
}
return result;
}
If you use Java 8 and would like to write less code you can use the Streams library.
List<Integer> list = Arrays.asList(1, 2, 3, 4, 5);
int[] array = list.stream().mapToInt(i -> i).toArray();
If you are open to using a third party library, you can Eclipse Collections as follows.
MutableList<Integer> list = Lists.mutable.with(1, 2, 3, 4, 5);
int[] array = list.collectInt(i -> i).toArray();
The following is slightly more code, but is the most efficient solution I could come up with using Eclipse Collections.
MutableList<Integer> list = Lists.mutable.with(1, 2, 3, 4, 5);
int[] array = new int[list.size()];
list.forEachWithIndex((each, index) -> array[index] = each);
If you need to use the java.util.List interface, the ListIterate utility class can be used from Eclipse Collections.
List<Integer> list = Arrays.asList(1, 2, 3, 4, 5);
int[] array = new int[list.size()];
ListIterate.forEachWithIndex(list, (each, index) -> array[index] = each);
The ListIterate utility will use different iteration code for RandomAccess lists and non-RandomAccess lists.
The most efficient thing to do would be to change the List<Integer> to a MutableIntList in Eclipse Collections or another library that has support for primitive collections.
Note: I am a committer for Eclipse Collections.
In Java 8:
int[] anArray = list.stream()
.filter(Objects::nonNull)
.mapToInt(Integer::intValue)
.toArray();
There is efficient way you could do this Java. However, this could open room for someone to create the generic function (depend on demand).
Just like this sample i wrote, I suggest you do the same to the specific knowledge of your program.
// Generic function to convert set to list
public static <T> ArrayList<T> convertSetToList(Set<T> set)
{
// create an empty list
ArrayList<T> list = new ArrayList<>();
// push each element in the set into the list
for (T t : set)
list.add(t);
// return the list
return list;
}
Is there any simple way to remove duplicate elements in Java(will two loops work and how). Thank you:)
IN: int[] arr = {1,3,4,2,3,1,6,7,7};
Output i want is:
{1,3,4,2,6,7}
the only i know is we can traverse it through loop.
eg.
for(int i = 0;i < arr.length;i++){
for(int j = 0;j<arr.length;j++){
if( ){
//what logic i can apply here.
}
}
}
This should work..
final Integer[] noDuplicates =
new LinkedHashSet<>(Arrays.asList(arr)).toArray(new Integer[0]);
Java 8 provides a nice way to do this using IntStream.
arr = Arrays.stream(arr) // Convert arr to IntStream
.distinct() // Get distinct elements (preserves order)
.toArray(); // Convert back to array
You can try this way:
int[]arr = {1,3,4,2,3,1,6,7,7};
Set <Integer> set = new HashSet<Integer>();
for ( int w: arr) {
set.add(w);
}
i think this would be helpful
public static int[] removeduplicate(int a[])
{
int count=0;
for(int i=0;i<a.length;i++)
{
for(int j=i+1;j<a.length;j++)
{
if(a[i]==a[j])
{
a[j]=-1;
count++;
}
}
}
int b[]=new int[a.length-count];
for(int i=0;i<a.length;i++)
{
if(a[i]!=-1)
{
for(int j=0;j<(a.length-count);j++)
{
b[j]=a[i];
}
}
}
return b;
}
You can use set data structure for removing duplicate elements from an array.
int arr[] = {1, 6, 4, 7, 8, 4, 7, 9, 3, 8};
Set<Integer> uniqueNumber = new HashSet<>();
for(int i = 0; i < arr.length; i++) {
//It only contains unique element
uniqueNumber.add(arr[i]);
}
For reference you can check this video tutorial https://www.youtube.com/watch?v=0HBIMjwte7s. I find it very helpful.
First u cant change the existing array , so need a new array to hold his unique numbers, again as you dont know the number of element in advance better to use ArrayList rather array. If you dont want to write to much logic or less number of loops , u can try this.
int[] arr = {1,3,4,2,3,1,6,7,7};
HashSet<Integer> hs = new HashSet<Integer>();
ArrayList<Integer> unique=new ArrayList<Integer>();
for(int num:arr){
if(hs.add(num)){
unique.add(num);
}
}
Is there any EASY way to sort an array in descending order like how they have a sort in ascending order in the Arrays class?
Or do I have to stop being lazy and do this myself :[
You could use this to sort all kind of Objects
sort(T[] a, Comparator<? super T> c)
Arrays.sort(a, Collections.reverseOrder());
Arrays.sort() cannot be used directly to sort primitive arrays in descending order. If you try to call the Arrays.sort() method by passing reverse Comparator defined by Collections.reverseOrder() , it will throw the error
no suitable method found for sort(int[],comparator)
That will work fine with 'Array of Objects' such as Integer array but will not work with a primitive array such as int array.
The only way to sort a primitive array in descending order is, first sort the array in ascending order and then reverse the array in place. This is also true for two-dimensional primitive arrays.
for a list
Collections.sort(list, Collections.reverseOrder());
for an array
Arrays.sort(array, Collections.reverseOrder());
You can use this:
Arrays.sort(data, Collections.reverseOrder());
Collections.reverseOrder() returns a Comparator using the inverse natural order. You can get an inverted version of your own comparator using Collections.reverseOrder(myComparator).
an alternative could be (for numbers!!!)
multiply the Array by -1
sort
multiply once again with -1
Literally spoken:
array = -Arrays.sort(-array)
without explicit comparator:
Collections.sort(list, Collections.reverseOrder());
with explicit comparator:
Collections.sort(list, Collections.reverseOrder(new Comparator()));
It's not directly possible to reverse sort an array of primitives (i.e., int[] arr = {1, 2, 3};) using Arrays.sort() and Collections.reverseOrder() because those methods require reference types (Integer) instead of primitive types (int).
However, we can use Java 8 Stream to first box the array to sort in reverse order:
// an array of ints
int[] arr = {1, 2, 3, 4, 5, 6};
// an array of reverse sorted ints
int[] arrDesc = Arrays.stream(arr).boxed()
.sorted(Collections.reverseOrder())
.mapToInt(Integer::intValue)
.toArray();
System.out.println(Arrays.toString(arrDesc)); // outputs [6, 5, 4, 3, 2, 1]
First you need to sort your array using:
Collections.sort(myArray);
Then you need to reverse the order from ascending to descending using:
Collections.reverse(myArray);
Java 8:
Arrays.sort(list, comparator.reversed());
Update:
reversed() reverses the specified comparator. Usually, comparators order ascending, so this changes the order to descending.
For array which contains elements of primitives if there is org.apache.commons.lang(3) at disposal easy way to reverse array (after sorting it) is to use:
ArrayUtils.reverse(array);
When an array is a type of Integer class then you can use below:
Integer[] arr = {7, 10, 4, 3, 20, 15};
Arrays.sort(arr, Collections.reverseOrder());
When an array is a type of int data type then you can use below:
int[] arr = {7, 10, 4, 3, 20, 15};
int[] reverseArr = IntStream.rangeClosed(1, arr.length).map(i -> arr[arr.length-i]).toArray();
I don't know what your use case was, however in addition to other answers here another (lazy) option is to still sort in ascending order as you indicate but then iterate in reverse order instead.
For discussions above, here is an easy example to sort the primitive arrays in descending order.
import java.util.Arrays;
public class Main {
public static void main(String[] args) {
int[] nums = { 5, 4, 1, 2, 9, 7, 3, 8, 6, 0 };
Arrays.sort(nums);
// reverse the array, just like dumping the array!
// swap(1st, 1st-last) <= 1st: 0, 1st-last: nums.length - 1
// swap(2nd, 2nd-last) <= 2nd: i++, 2nd-last: j--
// swap(3rd, 3rd-last) <= 3rd: i++, 3rd-last: j--
//
for (int i = 0, j = nums.length - 1, tmp; i < j; i++, j--) {
tmp = nums[i];
nums[i] = nums[j];
nums[j] = tmp;
}
// dump the array (for Java 4/5/6/7/8/9)
for (int i = 0; i < nums.length; i++) {
System.out.println("nums[" + i + "] = " + nums[i]);
}
}
}
Output:
nums[0] = 9
nums[1] = 8
nums[2] = 7
nums[3] = 6
nums[4] = 5
nums[5] = 4
nums[6] = 3
nums[7] = 2
nums[8] = 1
nums[9] = 0
Another solution is that if you're making use of the Comparable interface you can switch the output values which you had specified in your compareTo(Object bCompared).
For Example :
public int compareTo(freq arg0)
{
int ret=0;
if(this.magnitude>arg0.magnitude)
ret= 1;
else if (this.magnitude==arg0.magnitude)
ret= 0;
else if (this.magnitude<arg0.magnitude)
ret= -1;
return ret;
}
Where magnitude is an attribute with datatype double in my program. This was sorting my defined class freq in reverse order by it's magnitude. So in order to correct that, you switch the values returned by the < and >. This gives you the following :
public int compareTo(freq arg0)
{
int ret=0;
if(this.magnitude>arg0.magnitude)
ret= -1;
else if (this.magnitude==arg0.magnitude)
ret= 0;
else if (this.magnitude<arg0.magnitude)
ret= 1;
return ret;
}
To make use of this compareTo, we simply call Arrays.sort(mFreq) which will give you the sorted array freq [] mFreq.
The beauty (in my opinion) of this solution is that it can be used to sort user defined classes, and even more than that sort them by a specific attribute. If implementation of a Comparable interface sounds daunting to you, I'd encourage you not to think that way, it actually isn't. This link on how to implement comparable made things much easier for me. Hoping persons can make use of this solution, and that your joy will even be comparable to mine.
For 2D arrays to sort in descending order you can just flip the positions of the parameters
int[][] array= {
{1, 5},
{13, 1},
{12, 100},
{12, 85}
};
Arrays.sort(array, (a, b) -> Integer.compare(a[1], b[1])); // for ascending order
Arrays.sort(array, (b, a) -> Integer.compare(a[1], b[1])); // for descending order
Output for descending
12, 100
12, 85
1, 5
13, 1
You could use stream operations (Collections.stream()) with Comparator.reverseOrder().
For example, say you have this collection:
List<String> items = new ArrayList<>();
items.add("item01");
items.add("item02");
items.add("item03");
items.add("item04");
items.add("item04");
To print the items in their "natural" order you could use the sorted() method (or leave it out and get the same result):
items.stream()
.sorted()
.forEach(item -> System.out.println(item));
Or to print them in descending (reverse) order, you could use the sorted method that takes a Comparator and reverse the order:
items.stream()
.sorted(Comparator.reverseOrder())
.forEach(item -> System.out.println(item));
Note this requires the collection to have implemented Comparable (as do Integer, String, etc.).
There is a lot of mess going on here - people suggest solutions for non-primitive values, try to implement some sorting algos from the ground, give solutions involving additional libraries, showing off some hacky ones etc. The answer to the original question is 50/50. For those who just want to copy/paste:
// our initial int[] array containing primitives
int[] arrOfPrimitives = new int[]{1,2,3,4,5,6};
// we have to convert it into array of Objects, using java's boxing
Integer[] arrOfObjects = new Integer[arrOfPrimitives.length];
for (int i = 0; i < arrOfPrimitives.length; i++)
arrOfObjects[i] = new Integer(arrOfPrimitives[i]);
// now when we have an array of Objects we can use that nice built-in method
Arrays.sort(arrOfObjects, Collections.reverseOrder());
arrOfObjects is {6,5,4,3,2,1} now. If you have an array of something other than ints - use the corresponding object instead of Integer.
Simple method to sort an int array descending:
private static int[] descendingArray(int[] array) {
Arrays.sort(array);
int[] descArray = new int[array.length];
for(int i=0; i<array.length; i++) {
descArray[i] = array[(array.length-1)-i];
}
return descArray;
}
Adding my answer in here for a couple of different scenarios
For an Array
Arrays.sort(a, Comparator.reverseOrder());
FWIW Lists
Lists.reverse(a);
Any and all Collections
Collections.reverse(a);
I know that this is a quite old thread, but here is an updated version for Integers and Java 8:
Arrays.sort(array, (o1, o2) -> o2 - o1);
Note that it is "o1 - o2" for the normal ascending order (or Comparator.comparingInt()).
This also works for any other kinds of Objects. Say:
Arrays.sort(array, (o1, o2) -> o2.getValue() - o1.getValue());
This worked for me:
package doublearraysort;
import java.util.Arrays;
import java.util.Collections;
public class Gpa {
public static void main(String[] args) {
// initializing unsorted double array
Double[] dArr = new Double[] {
new Double(3.2),
new Double(1.2),
new Double(4.7),
new Double(3.3),
new Double(4.6),
};
// print all the elements available in list
for (double number : dArr) {
System.out.println("GPA = " + number);
}
// sorting the array
Arrays.sort(dArr, Collections.reverseOrder());
// print all the elements available in list again
System.out.println("The sorted GPA Scores are:");
for (double number : dArr) {
System.out.println("GPA = " + number);
}
}
}
Output:
GPA = 3.2
GPA = 1.2
GPA = 4.7
GPA = 3.3
GPA = 4.6
The sorted GPA Scores are:
GPA = 4.7
GPA = 4.6
GPA = 3.3
GPA = 3.2
GPA = 1.2
public double[] sortArrayAlgorithm(double[] array) { //sort in descending order
for (int i = 0; i < array.length; i++) {
for (int j = 0; j < array.length; j++) {
if (array[i] >= array[j]) {
double x = array[i];
array[i] = array[j];
array[j] = x;
}
}
}
return array;
}
just use this method to sort an array of type double in descending order, you can use it to sort arrays of any other types(like int, float, and etc) just by changing the "return type", the "argument type" and the variable "x" type to the corresponding type. you can also change ">=" to "<=" in the if condition to make the order ascending.
Another way with Comparator
import java.util.Arrays;
import java.util.Comparator;
...
Integer[] aInt = {6,2,3,4,1,5,7,8,9,10};
Arrays.sort(aInt, Comparator.reverseOrder() );
It's good sometimes we practice over an example, here is a full one:
sortdesc.java
import java.util.Arrays;
import java.util.Collections;
class sortdesc{
public static void main(String[] args){
// int Array
Integer[] intArray=new Integer[]{
new Integer(15),
new Integer(9),
new Integer(16),
new Integer(2),
new Integer(30)};
// Sorting int Array in descending order
Arrays.sort(intArray,Collections.reverseOrder());
// Displaying elements of int Array
System.out.println("Int Array Elements in reverse order:");
for(int i=0;i<intArray.length;i++)
System.out.println(intArray[i]);
// String Array
String[] stringArray=new String[]{"FF","PP","AA","OO","DD"};
// Sorting String Array in descending order
Arrays.sort(stringArray,Collections.reverseOrder());
// Displaying elements of String Array
System.out.println("String Array Elements in reverse order:");
for(int i=0;i<stringArray.length;i++)
System.out.println(stringArray[i]);}}
compiling it...
javac sortdec.java
calling it...
java sortdesc
OUTPUT
Int Array Elements in reverse order:
30
16
15
9
2
String Array Elements in reverse order:
PP
OO
FF
DD
AA
If you want to try an alphanumeric array...
//replace this line:
String[] stringArray=new String[]{"FF","PP","AA","OO","DD"};
//with this:
String[] stringArray=new String[]{"10FF","20AA","50AA"};
you gonna get the OUTPUT as follow:
50AA
20AA
10FF
source
There is a way that might be a little bit longer, but it works fine. This is a method to sort an int array descendingly.
Hope that this will help someone ,,, some day:
public static int[] sortArray (int[] array) {
int [] sortedArray = new int[array.length];
for (int i = 0; i < sortedArray.length; i++) {
sortedArray[i] = array[i];
}
boolean flag = true;
int temp;
while (flag) {
flag = false;
for (int i = 0; i < sortedArray.length - 1; i++) {
if(sortedArray[i] < sortedArray[i+1]) {
temp = sortedArray[i];
sortedArray[i] = sortedArray[i+1];
sortedArray[i+1] = temp;
flag = true;
}
}
}
return sortedArray;
}
I had the below working solution
public static int[] sortArrayDesc(int[] intArray){
Arrays.sort(intArray); //sort intArray in Asc order
int[] sortedArray = new int[intArray.length]; //this array will hold the sorted values
int indexSortedArray = 0;
for(int i=intArray.length-1 ; i >= 0 ; i--){ //insert to sortedArray in reverse order
sortedArray[indexSortedArray ++] = intArray [i];
}
return sortedArray;
}
Here is how I sorted a primitive type int array.
int[] intArr = new int[] {9,4,1,7};
Arrays.sort(nums);
Collections.reverse(Arrays.asList(nums));
Result:
[1, 4, 7, 9]
I know many answers are here, but still thinks , none of them tried using core java.
And using collection api , you will end up wasting so much memory and reseduals.
here is a try with pure core concepts , and yes this may be better way if you are more concerned about memory footprints.
int[] elements = new int [] {10,999,999,-58,548,145,255,889,1,1,4,5555,0,-1,-52};
//int[] elements = null;
if(elements != null && elements.length >1)
{
int max = 0, index = 0;
for(int i =0;i<elements.length;i++)//find out what is Max
{
if(elements[i] > max)
{
max = elements[i];
index = i;
}
}
elements[index] = elements[0];//Swap the places
elements[0] = max;
for(int i =0;i < elements.length;i++)//loop over element
{
for(int j = i+1;j < elements.length;j++)//loop to compare the elements
{
if(elements[j] > elements[i])
{
max = elements[j];
elements[j] = elements[i];
elements[i] = max;
}
}
}
}//i ended up using three loops and 2 extra variables
System.out.println(Arrays.toString(elements));//if null it will print null
// still love to learn more, please advise if we can do it better.
Love to learn from you too !
I'm trying to convert an ArrayList containing Integer objects to primitive int[] with the following piece of code, but it is throwing compile time error. Is it possible to convert in Java?
List<Integer> x = new ArrayList<Integer>();
int[] n = (int[])x.toArray(int[x.size()]);
If you are using java-8 there's also another way to do this.
int[] arr = list.stream().mapToInt(i -> i).toArray();
What it does is:
getting a Stream<Integer> from the list
obtaining an IntStream by mapping each element to itself (identity function), unboxing the int value hold by each Integer object (done automatically since Java 5)
getting the array of int by calling toArray
You could also explicitly call intValue via a method reference, i.e:
int[] arr = list.stream().mapToInt(Integer::intValue).toArray();
It's also worth mentioning that you could get a NullPointerException if you have any null reference in the list. This could be easily avoided by adding a filtering condition to the stream pipeline like this:
//.filter(Objects::nonNull) also works
int[] arr = list.stream().filter(i -> i != null).mapToInt(i -> i).toArray();
Example:
List<Integer> list = Arrays.asList(1, 2, 3, 4);
int[] arr = list.stream().mapToInt(i -> i).toArray(); //[1, 2, 3, 4]
list.set(1, null); //[1, null, 3, 4]
arr = list.stream().filter(i -> i != null).mapToInt(i -> i).toArray(); //[1, 3, 4]
You can convert, but I don't think there's anything built in to do it automatically:
public static int[] convertIntegers(List<Integer> integers)
{
int[] ret = new int[integers.size()];
for (int i=0; i < ret.length; i++)
{
ret[i] = integers.get(i).intValue();
}
return ret;
}
(Note that this will throw a NullPointerException if either integers or any element within it is null.)
EDIT: As per comments, you may want to use the list iterator to avoid nasty costs with lists such as LinkedList:
public static int[] convertIntegers(List<Integer> integers)
{
int[] ret = new int[integers.size()];
Iterator<Integer> iterator = integers.iterator();
for (int i = 0; i < ret.length; i++)
{
ret[i] = iterator.next().intValue();
}
return ret;
}
Google Guava
Google Guava provides a neat way to do this by calling Ints.toArray.
List<Integer> list = ...;
int[] values = Ints.toArray(list);
Apache Commons has a ArrayUtils class, which has a method toPrimitive() that does exactly this.
import org.apache.commons.lang.ArrayUtils;
...
List<Integer> list = new ArrayList<Integer>();
list.add(new Integer(1));
list.add(new Integer(2));
int[] intArray = ArrayUtils.toPrimitive(list.toArray(new Integer[0]));
However, as Jon showed, it is pretty easy to do this by yourself instead of using external libraries.
I believe iterating using the List's iterator is a better idea, as list.get(i) can have poor performance depending on the List implementation:
private int[] buildIntArray(List<Integer> integers) {
int[] ints = new int[integers.size()];
int i = 0;
for (Integer n : integers) {
ints[i++] = n;
}
return ints;
}
Java 8:
int[] intArr = Arrays.stream(integerList).mapToInt(i->i).toArray();
Arrays.setAll()
List<Integer> x = new ArrayList<>(Arrays.asList(7, 9, 13));
int[] n = new int[x.size()];
Arrays.setAll(n, x::get);
System.out.println("Array of primitive ints: " + Arrays.toString(n));
Output:
Array of primitive ints: [7, 9, 13]
The same works for an array of long or double, but not for arrays of boolean, char, byte, short or float. If you’ve got a really huge list, there’s even a parallelSetAll method that you may use instead.
To me this is good and elgant enough that I wouldn’t want to get an external library nor use streams for it.
Documentation link: Arrays.setAll(int[], IntUnaryOperator)
using Dollar should be quite simple:
List<Integer> list = $(5).toList(); // the list 0, 1, 2, 3, 4
int[] array = $($(list).toArray()).toIntArray();
I'm planning to improve the DSL in order to remove the intermediate toArray() call
This works nice for me :)
Found at https://www.techiedelight.com/convert-list-integer-array-int/
import java.util.Arrays;
import java.util.List;
class ListUtil
{
// Program to convert list of integer to array of int in Java
public static void main(String args[])
{
List<Integer> list = Arrays.asList(1, 2, 3, 4, 5);
int[] primitive = list.stream()
.mapToInt(Integer::intValue)
.toArray();
System.out.println(Arrays.toString(primitive));
}
}
Arrays.setAll() will work for most scenarios:
Integer List to primitive int array:
public static int[] convert(final List<Integer> list)
{
final int[] out = new int[list.size()];
Arrays.setAll(out, list::get);
return out;
}
Integer List (made of Strings) to primitive int array:
public static int[] convert(final List<String> list)
{
final int[] out = new int[list.size()];
Arrays.setAll(out, i -> Integer.parseInt(list.get(i)));
return out;
}
Integer array to primitive int array:
public static int[] convert(final Integer[] array)
{
final int[] out = new int[array.length];
Arrays.setAll(out, i -> array[i]);
return out;
}
Primitive int array to Integer array:
public static Integer[] convert(final int[] array)
{
final Integer[] out = new Integer[array.length];
Arrays.setAll(out, i -> array[i]);
return out;
}
It bewilders me that we encourage one-off custom methods whenever a perfectly good, well used library like Apache Commons has solved the problem already. Though the solution is trivial if not absurd, it is irresponsible to encourage such a behavior due to long term maintenance and accessibility.
Just go with Apache Commons
Java 8
int[] array = list.stream().mapToInt(i->i).toArray();
OR
int[] array = list.stream().mapToInt(Integer::intValue).toArray();
If you're using Eclipse Collections, you can use the collectInt() method to switch from an object container to a primitive int container.
List<Integer> integers = new ArrayList<>(Arrays.asList(1, 2, 3, 4, 5));
MutableIntList intList =
ListAdapter.adapt(integers).collectInt(i -> i);
Assert.assertArrayEquals(new int[]{1, 2, 3, 4, 5}, intList.toArray());
If you can convert your ArrayList to a FastList, you can get rid of the adapter.
Assert.assertArrayEquals(
new int[]{1, 2, 3, 4, 5},
Lists.mutable.with(1, 2, 3, 4, 5)
.collectInt(i -> i).toArray());
Note: I am a committer for Eclipse collections.
You can simply copy it to an array:
int[] arr = new int[list.size()];
for(int i = 0; i < list.size(); i++) {
arr[i] = list.get(i);
}
Not too fancy; but, hey, it works...
Next lines you can find convertion from int[] -> List -> int[]
private static int[] convert(int[] arr) {
List<Integer> myList=new ArrayList<Integer>();
for(int number:arr){
myList.add(number);
}
}
int[] myArray=new int[myList.size()];
for(int i=0;i<myList.size();i++){
myArray[i]=myList.get(i);
}
return myArray;
}
This code segment is working for me, try this:
Integer[] arr = x.toArray(new Integer[x.size()]);
Worth to mention ArrayList should be declared like this:
ArrayList<Integer> list = new ArrayList<>();
A very simple one-line solution is:
Integer[] i = arrlist.stream().toArray(Integer[]::new);
List<Integer> list = new ArrayList<Integer>();
list.add(1);
list.add(2);
int[] result = null;
StringBuffer strBuffer = new StringBuffer();
for (Object o : list) {
strBuffer.append(o);
result = new int[] { Integer.parseInt(strBuffer.toString()) };
for (Integer i : result) {
System.out.println(i);
}
strBuffer.delete(0, strBuffer.length());
}
Integer[] arr = (Integer[]) x.toArray(new Integer[x.size()]);
access arr like normal int[].