Reverse Array Order

Question

I am trying to reverse the order of an Array in java.
What is the most efficient way to do so in O(n) with the least amount of memory used.
No need to answer with code, pseudo code will be fine.
Here is my thought process:

  create a new temp array //I think this is a waste of memory, 
                          //but I am not sure if there's a better way
 grab elements from the end of the original array -decrement this variable
 insert element in beginning of temp array -increment this variable
then make the original array point to the temp array? //I am not sure 
            //if I can do this in java; so let's say the 
            //original array is Object[] arr; and the temp array is 
            //Object[] temp. Can I do temp = arr; ?

Is there a better more efficient way to do this perhaps without using a temp array? and Lastly, assume that there are no nulls in the array, so everything can work. Thank you

Edit: no this is not homework.

consider swaping first and last items and then the second and second last items till you reach half the list... you will just need one temporary variable and will still go over the list once? — Osama Javed
– Osama Javed, Commented Apr 3, 2012 at 14:36
Just loop through the original array in reverse order, and create a new container to contain the new order inserts. it's O(n) — Churk
– Churk, Commented Apr 3, 2012 at 14:38

Louis Wasserman · Accepted Answer · 2012-04-03 16:15:06Z

56

If it's an Object array, then Collections.reverse(Arrays.asList(array)) will do the job with constant memory and linear time -- no temporary array required.

edited Apr 3, 2012 at 16:15

answered Apr 3, 2012 at 14:39

Louis Wasserman

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4 Comments

Ernest Friedman-Hill Over a year ago

+1 Indeed, since the OP now says this is not homework, this is a great answer.

eddyparkinson Over a year ago

Love the solution. Just confirmed that no temporary array is required see: ideone.com/api/embed.js/link/xLLTpl ... click "Clone" and then "Run"

Jean-Yves Over a year ago

Does not work, at least with Java 1.6: System.out.println( X[ 0 ] + " to " + X[ X.length - 1 ] ); Collections.reverse( Arrays.asList( X ) ); System.out.println( X[ 0 ] + " to " + X[ X.length - 1 ] ); prints: 2272.6270739116 to 186.704625250768 2272.6270739116 to 186.704625250768

Louis Wasserman Over a year ago

@Jean-Yves: What is the type of X? I strongly suspect it's not an object array, which I specified in my answer was necessary.

ArjunShankar · Accepted Answer · 2013-06-23 22:20:04Z

13

Use a single temp element.

int array[SIZE];
int temp;

for (int i = 0; i < SIZE/2; i++)
  {
     temp = array[i];
     array[i] = array[SIZE-1 - i];
     array[SIZE-1 - i] = temp;
  }

edited Jun 23, 2013 at 22:20

answered Apr 3, 2012 at 14:36

ArjunShankar

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Comments

Ernest Friedman-Hill · Accepted Answer · 2012-04-03 14:36:52Z

12

You don't need to use a temporary array; just step through the array from the beginning to half-way through, swapping the element at i for the element at array.length-i-1. Be sure the handle the middle element correctly (not hard to do, but do make sure.)

answered Apr 3, 2012 at 14:36

Ernest Friedman-Hill

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Comments

Ted Hopp · Accepted Answer · 2012-04-03 14:39:05Z

2

you can do it without needing a temp array

loop from the beginning (or end doesn't matter) to the middle of the array
swap element with element at (last element - index) (so 0 and size - 1, 1 and size - 2 etc)
you'll do something like this to swap:

    temp = a[i];
    a[i] = a[end-i];
    a[end-i] = temp;

repeat

edited Apr 3, 2012 at 14:39

Ted Hopp

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answered Apr 3, 2012 at 14:36

twain249

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Comments

SirGuy · Accepted Answer · 2012-04-03 14:37:17Z

0

Here's two solutions:

    loop to N/2
      swap each element at i with element at N - i

Another solution is (depending on your circumstances) fake reversing the array by indexing:

    GetValueAt(int i){return array[N - i];}

answered Apr 3, 2012 at 14:37

SirGuy

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Comments

Andy · Accepted Answer · 2022-07-26 14:40:02Z

-1

Lets consider we have Integer array arr[] - Integer array.

for(int i=0,int J=arr.length-1 ; i<j ; i++,j--)
{
    temp =a[i];
    a[i]=a[j];
    a[j]=temp;
 }

this algorithm has O(n/2) as we will perfrom n/2 swaps. and its space complexity is constant 1 , as we are taking on temp variable.

edited Jul 26, 2022 at 14:40

answered Apr 20, 2015 at 9:52

Andy

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Comments

mcfinnigan · Accepted Answer · 2012-04-03 14:37:09Z

-2

pseudocode, assuming 0-based-index arrays:

for i in range(0, len(array)/2):
     swap(array[i], array[(len(array)-1)-i])

answered Apr 3, 2012 at 14:37

mcfinnigan

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1 Comment

ceving Over a year ago

This does not look like Java.

Darshan · Accepted Answer · 2017-06-21 03:32:52Z

-2

You can Do this in just two steps

ArrayList<Element> YourTempElement= new ArrayList<Element>(mElements);
Collections.reverse(YourTempElement);

answered Jun 21, 2017 at 3:32

Darshan

11

3 Comments

Nathan Tuggy Over a year ago

Uses same methods as accepted answer, just less elegantly and with less of an explanation.

Darshan Over a year ago

there are no need of explanation bro it's little two step nd i'm not explainer.

Nathan Tuggy Over a year ago

Good answers on Stack Overflow explain things. The accepted answer does. If there's already a good answer that says the same thing you would, or if there's no way to write a good answer at all, there's no real point in adding an answer to the question: that just adds noise.

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