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javaarraysmethodsreverse
Little_Berry
Little_Berry

Reputation: 23

Java reverse array method

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  
}

Upvotes: 1

Views: 18578

Answers (4)

Youssef
Youssef

Reputation: 11

int[] array = {1, 2, 3, 4, 5, 6, 7, 8, 9};
    System.out.println("The original Array: ");
    for (int i = 0; i < array.length; i++) {
        System.out.print(array[i] + " ");
    }
    System.out.println();
        System.out.println("The Reverse Array is: ");
        for (int i = array.length - 1; i >= 0; i--) {
            System.out.print(array[i] + " ");
       }

Upvotes: 1

Mohamad
Mohamad

Reputation: 29

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

   }

Upvotes: 3

user3657302
user3657302

Reputation: 347

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. :)

Upvotes: 1

Mureinik
Mureinik

Reputation: 311228

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.
}

Upvotes: 7

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