Efficient way to split a number in Java

Question

I'm struggling with finding an efficient way to split the digits of a number into a list.

For example, for the number:

23464646237

and the

ArrayList <Integer> vector;

i want to insert in each position of the array, a digit from the number, therefore obtaining the following output:

vector = {2,3,4,6,4,6,4,6,2,3,7}

How can i do this in a clean and efficient way in Java ?

Thanks in advance

@MarsAtomic because i'm worried about efficiency here, not just the reading part — laker001, Mar 20 '15 at 23:43
You might want to read all those answers in detail, because there is some discussion of efficiency. Look in particular at the modulus solution. All of this munging of Strings is quick and dirty, but not efficient. — MarsAtomic, Mar 20 '15 at 23:45

Jean-François Savard · Accepted Answer · 2015-03-20T23:47:04.223

The parseInt is not needed,

for(int i = 0; i < numberAsString.length(); i++)
    vector.add((int)numberAsString.charAt(i) - 48);

This will retrieve the ASCII value of the number at the position, then subtract 48 which is the value of 0 in the table.

Number from -127 to 127 are cached so I believe this will be the most efficient method (note that I'm not 100% sure). But anyway, only tried to give a way saving the parseInt call as you asked for the most efficient.

bmavus · Answer 2 · 2015-03-20T23:56:48.433

2

First of all don't convert it to string. Because it will work faster with numeric values. So you can use:

long nr = 23464646237;
while (nr > 0)
{
    vector.add(nr % 10);
    nr /= 10;
}

edited Mar 20 '15 at 23:56

answered Mar 20 '15 at 23:43

bmavus

892
1
7
21

Last line of the loop should be simplified to `nr /= 10;` – user207421 Mar 20 '15 at 23:53
@bmavus This won't compile, you will have an error saying that 23464646237 is too large for an integer. Change your declaration to `long nr = 23464646237l;`. – Jean-François Savard Mar 21 '15 at 17:21

score 2 · Answer 3 · answered Mar 20 '15 at 23:44

2

You could do something like this:

List<Integer> vector = Arrays.asList(String.valueOf(number).split(""))
            .stream()
            .map(Integer::parseInt)
            .collect(Collectors.toList());

answered Mar 20 '15 at 23:44

kjosh

587
4
11

Theo · Answer 4 · 2015-03-20T23:50:43.593

2

Just make your number a string and make use of the charAt() method to add each char of the string into a List.

public static ArrayList<Object>vect=new ArrayList<Object>();


public static void main(String[] args){
    String y="123321312312";
    for (int i=0;i<y.length();i++){
        vect.add(y.charAt(i));
    }
}

edited Mar 20 '15 at 23:50

answered Mar 20 '15 at 23:44

Theo

1,165
9
14

Dermot Blair · Answer 5 · 2015-03-21T17:19:03.923

1

Try this:

long yourNumber = 23464646237L;
ArrayList<Integer> vector = new ArrayList<Integer>();
String numStr = String.valueOf(yourNumber);
for(int i=0;i<numStr.length();i++)
{
    vector.add(Integer.valueOf(numStr.charAt(i)));
}

edited Mar 21 '15 at 17:19

answered Mar 20 '15 at 23:40

Dermot Blair

1,600
10
10

I didnt downvote, but, `numStr` is not array to go by index. – RP- Mar 20 '15 at 23:54
Thanks. Im getting mixed up with C++ where you can use [] to access a character at an index of a string :) – Dermot Blair Mar 21 '15 at 00:00
@DermotBlair This won't compile, you will have an error saying that `23464646237 ` is too large for an integer. Change your declaration to `long yourNumber = 23464646237l;`. – Jean-François Savard Mar 21 '15 at 17:10
@Jean-FrançoisSavard its updated now, thanks. – Dermot Blair Mar 21 '15 at 17:20

Efficient way to split a number in Java

5 Answers5