I am trying to reverse an int array in Java.
This method does not reverse the array.
for(int i = 0; i < validData.length; i++)
{
int temp = validData[i];
validData[i] = validData[validData.length - i - 1];
validData[validData.length - i - 1] = temp;
}
What is wrong with it?
To reverse an int array, you swap items up until you reach the midpoint, like this:
for(int i = 0; i < validData.length / 2; i++)
{
int temp = validData[i];
validData[i] = validData[validData.length - i - 1];
validData[validData.length - i - 1] = temp;
}
The way you are doing it, you swap each element twice, so the result is the same as the initial list.
With Commons.Lang, you could simply use
ArrayUtils.reverse(int[] array)
Most of the time, it's quicker and more bug-safe to stick with easily available libraries already unit-tested and user-tested when they take care of your problem.
private int[] reversed(int[] array) {ArrayUtils.reverse(array);}
and then you can do f(reversed(arr));
:))
Collections.reverse(Arrays.asList(yourArray));
java.util.Collections.reverse()
can reverse java.util.List
s and java.util.Arrays.asList()
returns a list that wraps the the specific array you pass to it, therefore yourArray
is reversed after the invocation of Collections.reverse()
.
The cost is just the creation of one List-object and no additional libraries are required.
A similar solution has been presented in the answer of Tarik and their commentors, but I think this answer would be more concise and more easily parsable.
int[]
to asList(...)
will not return a List<Integer>
, but a List<int[]>
, containing one element. There is AFAICS no simple built-in way to convert an int[]
to an Integer[]
.
Arrays.stream(arr).boxed().collect(Collectors.toList())
or Arrays.stream(arr).boxed().toArray(Integer[]::new)
public class ArrayHandle {
public static Object[] reverse(Object[] arr) {
List<Object> list = Arrays.asList(arr);
Collections.reverse(list);
return list.toArray();
}
}
int
s in this case) are wrapped in to their respective wrappers (Integer
s in this case) and put in the list. You see, Integer
s are objects. @Tom
int[]
is different from Integer[]
. Try it: Integer[] array = new int[5]
. You'll get a compile error. This is why the Java Arrays
class defines a bunch of methods for working with primitive arrays. Trying to pass an int[]
to the above method will result in something like The method reverse(Object[]) in the type MakeSimple is not applicable for the arguments (int[])
. @Filip - an in-place algorithm uses less memory and runs faster.
Arrays.asList()
doesn't reference the original array, nor does the array returned. That's one of the problems with this method: it uses triple the memory and does triple the work as an in-place algorithm.
int[]
as an argument to this method ("incompatible types: int[] cannot be converted to Object[]").
I think it's a little bit easier to follow the logic of the algorithm if you declare explicit variables to keep track of the indices that you're swapping at each iteration of the loop.
public static void reverse(int[] data) {
for (int left = 0, right = data.length - 1; left < right; left++, right--) {
// swap the values at the left and right indices
int temp = data[left];
data[left] = data[right];
data[right] = temp;
}
}
I also think it's more readable to do this in a while loop.
public static void reverse(int[] data) {
int left = 0;
int right = data.length - 1;
while( left < right ) {
// swap the values at the left and right indices
int temp = data[left];
data[left] = data[right];
data[right] = temp;
// move the left and right index pointers in toward the center
left++;
right--;
}
}
There are already a lot of answers here, mostly focused on modifying the array in-place. But for the sake of completeness, here is another approach using Java streams to preserve the original array and create a new reversed array:
int[] a = {8, 6, 7, 5, 3, 0, 9};
int[] b = IntStream.rangeClosed(1, a.length).map(i -> a[a.length-i]).toArray();
In case of Java 8 we can also use IntStream
to reverse the array of integers as:
int[] sample = new int[]{1,2,3,4,5};
int size = sample.length;
int[] reverseSample = IntStream.range(0,size).map(i -> sample[size-i-1])
.toArray(); //Output: [5, 4, 3, 2, 1]
With Guava:
Collections.reverse(Ints.asList(array));
asList
methods, it creates a view that writes directly through to the backing (primitive) array. I think the down-voter here mistakenly thought that this returned a boxed list or something.
Collections.reverse
is a void method. This operates in-place on a Guava internal class which wraps an int[]
(Since it never stores a list of boxed Integer
s I wouldn't call the class a "boxed list", but rather a "List view of an array"). But yes it operates via an interface passing Integer
objects, so this would create a lot of temporary object churn and boxing as mentioned. Try an IntStream
or a primitive-collection library for where performance matters. (Trove, Koloboke, Eclipse Collections, ...)
for(int i=validData.length-1; i>=0; i--){
System.out.println(validData[i]);
}
Simple for loop!
for (int start = 0, end = array.length - 1; start <= end; start++, end--) {
int aux = array[start];
array[start]=array[end];
array[end]=aux;
}
start <= end
to start < end
This will help you
int a[] = {1,2,3,4,5};
for (int k = 0; k < a.length/2; k++) {
int temp = a[k];
a[k] = a[a.length-(1+k)];
a[a.length-(1+k)] = temp;
}
This is how I would personally solve it. The reason behind creating the parametrized method is to allow any array to be sorted... not just your integers.
I hope you glean something from it.
@Test
public void reverseTest(){
Integer[] ints = { 1, 2, 3, 4 };
Integer[] reversedInts = reverse(ints);
assert ints[0].equals(reversedInts[3]);
assert ints[1].equals(reversedInts[2]);
assert ints[2].equals(reversedInts[1]);
assert ints[3].equals(reversedInts[0]);
reverseInPlace(reversedInts);
assert ints[0].equals(reversedInts[0]);
}
@SuppressWarnings("unchecked")
private static <T> T[] reverse(T[] array) {
if (array == null) {
return (T[]) new ArrayList<T>().toArray();
}
List<T> copyOfArray = Arrays.asList(Arrays.copyOf(array, array.length));
Collections.reverse(copyOfArray);
return copyOfArray.toArray(array);
}
private static <T> T[] reverseInPlace(T[] array) {
if(array == null) {
// didn't want two unchecked suppressions
return reverse(array);
}
Collections.reverse(Arrays.asList(array));
return array;
}
Collections.reverse(asList(arraytoReverse)); return arrayToReverse;
. asList
is just a wrapper around the array, so the original array is reversed.
If working with data that is more primitive (i.e. char, byte, int, etc) then you can do some fun XOR operations.
public static void reverseArray4(int[] array) {
int len = array.length;
for (int i = 0; i < len/2; i++) {
array[i] = array[i] ^ array[len - i - 1];
array[len - i - 1] = array[i] ^ array[len - i - 1];
array[i] = array[i] ^ array[len - i - 1];
}
}
Your program will work for only length = 0, 1
. You can try :
int i = 0, j = validData.length-1 ;
while(i < j)
{
swap(validData, i++, j--); // code for swap not shown, but easy enough
}
There are two ways to have a solution for the problem:
1. Reverse an array in space.
Step 1. Swap the elements at the start and the end index.
Step 2. Increment the start index decrement the end index.
Step 3. Iterate Step 1 and Step 2 till start index < end index
For this, the time complexity will be O(n) and the space complexity will be O(1)
Sample code for reversing an array in space is like:
public static int[] reverseAnArrayInSpace(int[] array) {
int startIndex = 0;
int endIndex = array.length - 1;
while(startIndex < endIndex) {
int temp = array[endIndex];
array[endIndex] = array[startIndex];
array[startIndex] = temp;
startIndex++;
endIndex--;
}
return array;
}
2. Reverse an array using an auxiliary array.
Step 1. Create a new array of size equal to the given array.
Step 2. Insert elements to the new array starting from the start index, from the given array starting from end index.
For this, the time complexity will be O(n) and the space complexity will be O(n)
Sample code for reversing an array with auxiliary array is like:
public static int[] reverseAnArrayWithAuxiliaryArray(int[] array) {
int[] reversedArray = new int[array.length];
for(int index = 0; index < array.length; index++) {
reversedArray[index] = array[array.length - index -1];
}
return reversedArray;
}
Also, we can use the Collections API from Java to do this.
The Collections API internally uses the same reverse in space approach.
Sample code for using the Collections API is like:
public static Integer[] reverseAnArrayWithCollections(Integer[] array) {
List<Integer> arrayList = Arrays.asList(array);
Collections.reverse(arrayList);
return arrayList.toArray(array);
}
There are some great answers above, but this is how I did it:
public static int[] test(int[] arr) {
int[] output = arr.clone();
for (int i = arr.length - 1; i > -1; i--) {
output[i] = arr[arr.length - i - 1];
}
return output;
}
It is most efficient to simply iterate the array backwards.
I'm not sure if Aaron's solution does this vi this call Collections.reverse(list);
Does anyone know?
public void getDSCSort(int[] data){
for (int left = 0, right = data.length - 1; left < right; left++, right--){
// swap the values at the left and right indices
int temp = data[left];
data[left] = data[right];
data[right] = temp;
}
}
Solution with o(n) time complexity and o(1) space complexity.
void reverse(int[] array) {
int start = 0;
int end = array.length - 1;
while (start < end) {
int temp = array[start];
array[start] = array[end];
array[end] = temp;
start++;
end--;
}
}
for (int start = 0, end = array.length - 1; start < end; start++, end--) { ... }
.
public void display(){
String x[]=new String [5];
for(int i = 4 ; i > = 0 ; i-- ){//runs backwards
//i is the nums running backwards therefore its printing from
//highest element to the lowest(ie the back of the array to the front) as i decrements
System.out.println(x[i]);
}
}
Wouldn't doing it this way be much more unlikely for mistakes?
int[] intArray = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
int[] temp = new int[intArray.length];
for(int i = intArray.length - 1; i > -1; i --){
temp[intArray.length - i -1] = intArray[i];
}
intArray = temp;
Using the XOR solution to avoid the temp variable your code should look like
for(int i = 0; i < validData.length; i++){
validData[i] = validData[i] ^ validData[validData.length - i - 1];
validData[validData.length - i - 1] = validData[i] ^ validData[validData.length - i - 1];
validData[i] = validData[i] ^ validData[validData.length - i - 1];
}
See this link for a better explanation:
http://betterexplained.com/articles/swap-two-variables-using-xor/
2 ways to reverse an Array .
Using For loop and swap the elements till the mid point with time complexity of O(n/2). private static void reverseArray() { int[] array = new int[] { 1, 2, 3, 4, 5, 6 }; for (int i = 0; i < array.length / 2; i++) { int temp = array[i]; int index = array.length - i - 1; array[i] = array[index]; array[index] = temp; } System.out.println(Arrays.toString(array)); } Using built in function (Collections.reverse()) private static void reverseArrayUsingBuiltInFun() { int[] array = new int[] { 1, 2, 3, 4, 5, 6 }; Collections.reverse(Ints.asList(array)); System.out.println(Arrays.toString(array)); } Output : [6, 5, 4, 3, 2, 1]
Ints
?
public static void main(String args[]) {
int [] arr = {10, 20, 30, 40, 50};
reverse(arr, arr.length);
}
private static void reverse(int[] arr, int length) {
for(int i=length;i>0;i--) {
System.out.println(arr[i-1]);
}
}
below is the complete program to run in your machine.
public class ReverseArray {
public static void main(String[] args) {
int arr[] = new int[] { 10,20,30,50,70 };
System.out.println("reversing an array:");
for(int i = 0; i < arr.length / 2; i++){
int temp = arr[i];
arr[i] = arr[arr.length - i - 1];
arr[arr.length - i - 1] = temp;
}
for (int i = 0; i < arr.length; i++) {
System.out.println(arr[i]);
}
}
}
For programs on matrix using arrays this will be the good source.Go through the link.
private static int[] reverse(int[] array){
int[] reversedArray = new int[array.length];
for(int i = 0; i < array.length; i++){
reversedArray[i] = array[array.length - i - 1];
}
return reversedArray;
}
Here is a simple implementation, to reverse array of any type, plus full/partial support.
import java.util.logging.Logger;
public final class ArrayReverser {
private static final Logger LOGGER = Logger.getLogger(ArrayReverser.class.getName());
private ArrayReverser () {
}
public static <T> void reverse(T[] seed) {
reverse(seed, 0, seed.length);
}
public static <T> void reverse(T[] seed, int startIndexInclusive, int endIndexExclusive) {
if (seed == null || seed.length == 0) {
LOGGER.warning("Nothing to rotate");
}
int start = startIndexInclusive < 0 ? 0 : startIndexInclusive;
int end = Math.min(seed.length, endIndexExclusive) - 1;
while (start < end) {
swap(seed, start, end);
start++;
end--;
}
}
private static <T> void swap(T[] seed, int start, int end) {
T temp = seed[start];
seed[start] = seed[end];
seed[end] = temp;
}
}
Here is the corresponding Unit Test
import static org.hamcrest.CoreMatchers.is;
import static org.junit.Assert.assertThat;
import org.junit.Before;
import org.junit.Test;
public class ArrayReverserTest {
private Integer[] seed;
@Before
public void doBeforeEachTestCase() {
this.seed = new Integer[]{1,2,3,4,5,6,7,8};
}
@Test
public void wholeArrayReverse() {
ArrayReverser.<Integer>reverse(seed);
assertThat(seed[0], is(8));
}
@Test
public void partialArrayReverse() {
ArrayReverser.<Integer>reverse(seed, 1, 5);
assertThat(seed[1], is(5));
}
}
Here is what I've come up with:
// solution 1 - boiler plated
Integer[] original = {100, 200, 300, 400};
Integer[] reverse = new Integer[original.length];
int lastIdx = original.length -1;
int startIdx = 0;
for (int endIdx = lastIdx; endIdx >= 0; endIdx--, startIdx++)
reverse[startIdx] = original[endIdx];
System.out.printf("reverse form: %s", Arrays.toString(reverse));
// solution 2 - abstracted
// convert to list then use Collections static reverse()
List<Integer> l = Arrays.asList(original);
Collections.reverse(l);
System.out.printf("reverse form: %s", l);
static int[] reverseArray(int[] a) {
int ret[] = new int[a.length];
for(int i=0, j=a.length-1; i<a.length && j>=0; i++, j--)
ret[i] = a[j];
return ret;
}
public static int[] reverse(int[] array) {
int j = array.length-1;
// swap the values at the left and right indices //////
for(int i=0; i<=j; i++)
{
int temp = array[i];
array[i] = array[j];
array[j] = temp;
j--;
}
return array;
}
public static void main(String []args){
int[] data = {1,2,3,4,5,6,7,8,9};
reverse(data);
}
Success story sharing
validData.length / 2
part to the outside of the for-loop.validData.length >> 1
. That is equivalent and faster, but it confuses many programmers and any good compiler will automatically do that.validData.length - i - 1
and save it to a variable.