CODEWITHSUNDEEP
javaarraysinteger
Martin Tuskevicius
Martin Tuskevicius

Reputation: 2630

Java; int[] to int

How would I convert an array of integers into an integer, but in a special way?

For example, how would I convert { 1, 9, 9, 0 } into 1990?

Upvotes: 3

Views: 53256

Answers (7)

Andy Turner
Andy Turner

Reputation: 140534

It is upsetting to see the unnecessarily complex solutions, involving Math.pow, String etc. Even regex and third party libraries make an appearance!

All you need is a loop and multiplication by 10:

int num = 0;
for (int a : arr) {
  num = 10*num + a;
}

This of course assumes that the elements of arr are in range 0-9; but then the correct behavior outside that case is undefined anyway.

It also assumes that the array's digits can be converted to an integer without overflow; but since the question asks about converting to an integer, overflow behavior is also undefined.

Note that you can handle the "element > 9" case in the similar way to the string concatenation solution by working out how many digits you have to shift num by:

int num = 0;
for (int a : arr) {
  if (a < 0) throw new IllegalArgumentException();
  int i = a;
  do {
    num *= 10;
    i /= 10;
  } while (i != 0);
  num += a;
}

I don't think this is the most appropriate way to handle multiple digits in an element; at the very least the behaviour should be specified. I'm just pointing out that you don't need to resort to strings to handle it.

Upvotes: 4

Daniel Dil
Daniel Dil

Reputation: 51

Use:

public void turnArrToInt(int x[]){
    for(int i = 0; i < x.length; i++){
        number += x[i]*Math.pow(10,x.length-i-1);
    }

This is how I implemented it. It takes each element, and if it is the first of the array it multiplies it by 10^(length of the array - 1 - theCurrentElement) and it adds all to the result. It works for all sizes. Basically multiplying the 100th by 100, 10th by 10 and so on.

For example: char = {1,2,3} would multiply 1 by 100 2 by 10 3 by 1 and get the sum of all and get 123.

Upvotes: 0

davecoulter
davecoulter

Reputation: 1826

I am not sure in Java, but in pseudo .NET code:

String value = "";
for (int i = 0; i < array.length; i++)
{
    value += array[i]; // Build out the number as a string
}

int someInt = Integer.parseInt(value);

Upvotes: 1

rednoyz
rednoyz

Reputation: 1327

Arrays.toString(nums).replaceAll("\\D+","");

Upvotes: -1

Prashant Bhate
Prashant Bhate

Reputation: 11087

Assuming you have digits in Integer[] instead of primitive int[] and have Commons Lang Library you may find following one liner useful

    Integer[] array = {1, 9, 9, 0 };
    System.out.println(Integer.valueOf(StringUtils.join(array)));

Or if the integer is too big to fit in int use BigInteger 

    Integer[] piDigits = { 3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9, 3, 2,
            3, 8, 4, 6, 2, 6, 4, 3, 3, 8, 3, 2, 7, 9, 5, 0, 2, 8, 8, 4, 1,
            9, 7, 1, 6, 9, 3, 9, 9, 3, 7, 5, 1, 0, 5, 8, 2, 0, 9, 7, 4, 9,
            4, 4, 5, 9, 2, 3, 0, 7, 8, 1, 6, 4, 0, 6, 2, 8, 6 };
    System.out.println(new BigInteger(StringUtils.join(piDigits)));

Upvotes: 1

Shawn
Shawn

Reputation: 7365

This will work for any size integer array

int nums[] = { 1, 9, 9, 0 };

StringBuilder strNum = new StringBuilder();

for (int num : nums) 
{
     strNum.append(num);
}
int finalInt = Integer.parseInt(strNum.toString());
System.out.println(finalInt);

Upvotes: 20

Eng.Fouad
Eng.Fouad

Reputation: 117675

int[] array = {1,9,9,0};
int result = 0;
for(int i = 0; i < array.length; i++) result += Math.pow(10,i) * array[array.length - i - 1];
System.out.println(result);

Output: 1990

Upvotes: 7

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