current community

your communities

more stack exchange communities

Stack Exchange
tour help
stack overflow careers
Take the 2-minute tour ×
Stack Overflow is a question and answer site for professional and enthusiast programmers. It's 100% free.
up vote 1 down vote favorite

I'm trying to create a method that takes in an array and then returns that array in reverse. The code I wrote returns the array in reverse, but, the first two values are now 0. Anyone know what I did wrong?

public static int[] reverse(int[] x)
{     
    int []d = new int[x.length];

    for (int i = 0; i < x.length/2; i++)  // for loop, that checks each array slot
    {
        d[i] = x[i];
        x[i] = x[x.length-1-i];  // creates a new array that is in reverse order of the original
        x[x.length-1-i] = d[i];
    }
    return d;      // returns the new reversed array  
}
share|improve this question
    
Maybe you should rethink your solution and try your algorithm on paper before you write some code.. –  Billda Jul 1 '14 at 19:53
    
Note that you're directly writing in x when you should use it to read the data only. –  Luiggi Mendoza Jul 1 '14 at 19:54
1  
@Eran because you don't need to iterate over the whole array to access every element. Note how OP goes for the array data: x[i] and x[x.length - 1 - i]. –  Luiggi Mendoza Jul 1 '14 at 19:59
2  
It looks to me like x is reversed in place, I don't see why d would be returned. –  Andreas Jul 1 '14 at 19:59

3 Answers 3

up vote 5 down vote

You are assigning values from an uninitialized array d to x - that's where the zeroes (default value for an int in Java) are coming from.

IIUC, you're mixing two reversing strategies.

If you're creating a new array, you needn't run over half of the original array, but over all of it:

public static int[] reverse(int[] x) {

    int[] d = new int[x.length];


    for (int i = 0; i < x.length; i++) {
        d[i] = x[x.length - 1 -i];
    }
    return d;
}

Alternatively, if you want to reverse the array in place, you don't need a temp array, only a single variable (at most - there are also ways to switch two ints without an additional variable, but that's a different question):

public static int[] reverseInPlace(int[] x) {
    int tmp;    

    for (int i = 0; i < x.length / 2; i++) {
        tmp = x[i];
        x[i] = x[x.length - 1 - i];
        x[x.length - 1 - i] = tmp;
    }
    return x; // for completeness, not really necessary.
}
share|improve this answer
    
I changed it to you first suggestions, but what happens is that when it prints, the first and last values are switched around, any idea why? –  Diz Jul 1 '14 at 21:49
up vote 1 down vote

Here is a short way to do it.

  public static int[] reverse(int[] x)
   {

       int[] d = new int[x.length];            //create new array


       for (int i=x.length-1; i < 0; i--)      // revered loop
       {
        d[(x.length-i-1)]=x[i];                 //setting values              

       }
        return d[];                            // returns the new reversed array

   }
share|improve this answer
up vote 0 down vote

Its simple mistake; you are coping reversed data in x; and returning d. If you will return x, you will get complete revered data.

    d[i] = x[i];    // you are copying first element to some temp value
    x[i] = x[x.length-1-i];  // copied last element to first; and respective...
    x[x.length-1-i] = d[i]; // copied temp element to first element; and temp elements are nothing but array d

So ultimately you have created revered array inside x and not in d. If you will return x you got your answer. And d which is just half baked; so you get default value of 0 for remainign half array. :)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.