Creating an array of indeterminate length?

Question

I'm supposed to add two numbers, in the form of an array,

example

839 as an array, {9,3,8}

2039 {9,3,0,2}

The reason it is backward is because we're required to start from the one's place. I just can't figure what the size of the array should be.

public static int[] add(int [] x, int[] y){

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

     int j=0,
         hold,
         i;

     for(i=0; i<=z.length-1;i++) {

        z[i]=(x[i]+y[i]+j)%10;
        j=(x[i]+y[i]+j)/10;

     }

This is what I have currently, but the size of the array is the problem. I feel like I've tried everything but idk.

You can have the size of the array as the maximum length of the two numbers. — theboringdeveloper, Apr 18 '18 at 04:42
The array isn't big enough for all of the numbers so if I'm adding numbers of different lengths, some numbers get cut off, and if they're the same it'll have extra 0's in front of it. — Fluxia, Apr 18 '18 at 04:43
Possible duplicate of [Java dynamic array sizes?](https://stackoverflow.com/questions/1647260/java-dynamic-array-sizes) — sshanzel, Apr 18 '18 at 04:44

score 0 · Answer 1 · answered Apr 18 '18 at 04:46

It's not actually possible to determine the exact size of the result without performing the addition; all you really know for sure upfront is that it will be either Math.max(x.length, y.length) or Math.max(x.length, y.length) + 1.

So, I think there are three approaches:

You can always allocate Math.max(x.length, y.length) + 1, and accept that sometimes there will be an extra leading zero.
- But in this case, in order to avoid adding more and more zeroes with every single addition, you'll probably want to pre-count the number of leading zeroes in x and y, and discount those from your total. This ends up being unnecessarily complicated IMHO.
You can do the addition twice: once to determine the length so you can create the array, and once to actually populate it with the result.
You can initially allocate Math.max(x.length, y.length) + 1 and do the addition, and then if the most-significant-digit turned out to be zero, you can then allocate Math.max(x.length, y.length) and copy all but that zero. (You may want to use java.lang.System.arraycopy(...) for this.)

Unfortunately my professor is very anal and isn't letting us us Math. Otherwise I would've been done already, is there another way to do it? — Fluxia, Apr 18 '18 at 04:49
@Fluxia: I'm not saying you need to use `Math`. I know Java, and you know Java, so `Math.max(x.length, y.length)` is just a convenient way for me to tell you "either `x.length` or `y.length`, whichever is greater". It's trivial for you to implement that using an `if` statement, or using the ternary operator `... ? ... : ...` if you're familiar with it. — ruakh, Apr 18 '18 at 04:51

score 0 · Answer 2 · edited Apr 18 '18 at 04:53

0

If you use an Integer object instead of a primitive int you can do the following.

Integer x = 1234;
x.toString().length();

edited Apr 18 '18 at 04:53

4b0

21,981
30
95
142

answered Apr 18 '18 at 04:47

Bilal Shehata

51
8

I tried this already and my professor failed me on my project initially. Is there any other way to do it? – Fluxia Apr 18 '18 at 04:50
you can use Arrays.copyOf and then repeatedly create a new array of the current size + 1 (as needed) – Bilal Shehata Apr 18 '18 at 04:55

Oleg Cherednik · Answer 3 · 2018-04-18T08:38:26.773

What about this one?

public static int[] add(int[] x, int[] y) {
    int[] z = new int[x.length + y.length];
    int i = 0;

    for (int j = 0; i < x.length || i < y.length || j > 0; i++) {
        int sum = (i < x.length ? x[i] : 0) + (i < y.length ? y[i] : 0) + j;
        z[i] = sum % 10;
        j = sum / 10;
    }

//    return Arrays.copyOf(z, i);

    int[] res = new int[i];
    System.arraycopy(z, 0, res, 0, res.length);

    return res;
}

Creating an array of indeterminate length?

3 Answers3