Gets last digit of a number

Question

I need to define the last digit of a number assign this to value. After this, return the last digit.

My snippet of code doesn't work correctly...

Code:

public int lastDigit(int number) {
    String temp = Integer.toString(number);
    int[] guess = new int[temp.length()];
    int last = guess[temp.length() - 1];

    return last;
}

Question:

How to solve this issue?

Why does it not work? Do you get the wrong value or an exception? — Tomas McGuinness, Jun 17 '13 at 10:09
You're creating an empty array when you create "guess". You need to populate it with the characters of temp too. The right way would be to use %10 as most people have answered below. But to fix your code without changing the approach, you can do something like `Integer.parseInt(temp.substring(temp.length()-1))` — Rajesh J Advani, Jun 17 '13 at 10:31

Bathsheba · Accepted Answer · 2014-05-09T14:50:51.890

168

Just return (number % 10); i.e. take the modulus. This will be much faster than parsing in and out of a string.

If number can be negative then use (Math.abs(number) % 10);

edited May 09 '14 at 14:50

answered Jun 17 '13 at 10:10

Bathsheba

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What is Java's semantics of the % operator when the left operand is negative? Chances are that it will not return the last digit. – Kaz Apr 18 '14 at 21:42
@Kaz: That's an important point: %10 does not extract the last digit for a negative number. Thank you. I've amended. – Bathsheba May 09 '14 at 14:56
What if number is null? – Thermech Feb 01 '16 at 21:13
OP has a primitive type for number: so can't be null. – Bathsheba Feb 01 '16 at 22:18

score 18 · Answer 2 · edited Jun 04 '20 at 14:26

18

Below is a simpler solution how to get the last digit from an int:

public int lastDigit(int number) { return Math.abs(number) % 10; }

edited Jun 04 '20 at 14:26

SmushyTaco

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answered Jun 17 '13 at 10:10

Adam Siemion

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You can be first to get it right for when `number` is negative. – Kaz Apr 18 '14 at 21:43

score 8 · Answer 3 · answered Jun 17 '13 at 10:13

8

Use

int lastDigit = number % 10.

Read about Modulo operator: http://en.wikipedia.org/wiki/Modulo_operation

Or, if you want to go with your String solution

String charAtLastPosition = temp.charAt(temp.length()-1);

answered Jun 17 '13 at 10:13

darijan

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score 5 · Answer 4 · answered Jun 17 '13 at 10:11

5

No need to use any strings.Its over burden.

int i = 124;
int last= i%10;
System.out.println(last);   //prints 4

answered Jun 17 '13 at 10:11

Suresh Atta

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score 2 · Answer 5 · answered Dec 13 '15 at 18:28

2

Without using '%'.

public int lastDigit(int no){
    int n1 = no / 10;
    n1 = no - n1 * 10;
    return n1;
}

answered Dec 13 '15 at 18:28

What would be the point? I only see needless complication. – Ole V.V. Dec 02 '21 at 05:03

score 1 · Answer 6 · answered Jun 17 '13 at 10:10

1

You have just created an empty integer array. The array guess does not contain anything to my knowledge. The rest you should work out to get better.

answered Jun 17 '13 at 10:10

Peter Jaloveczki

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score 1 · Answer 7 · answered Jun 17 '13 at 10:48

1

Your array don't have initialization. So it will give default value Zero. You can try like this also

String temp = Integer.toString(urNumber);
System.out.println(temp.charAt(temp.length()-1));

answered Jun 17 '13 at 10:48

Sivaraman

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score 1 · Answer 8 · edited Mar 14 '19 at 22:50

1

public static void main(String[] args) {

    System.out.println(lastDigit(2347));
}

public static int lastDigit(int number)
{
    //your code goes here. 
    int last = number % 10;

    return last;
}

0/p:

7

edited Mar 14 '19 at 22:50

Samuel Liew

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answered Sep 24 '14 at 17:11

Manjunath

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score 0 · Answer 9 · edited Mar 14 '19 at 22:51

0

here is your method

public int lastDigit(int number)
{
    //your code goes here. 
    int last =number%10;
    return last;
}

edited Mar 14 '19 at 22:51

Samuel Liew

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answered Jun 17 '13 at 10:20

KhAn SaAb

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score 0 · Answer 10 · answered Jun 21 '17 at 07:43

0

Use StringUtils, in case you need string result:

String last = StringUtils.right(number.toString(), 1);

answered Jun 21 '17 at 07:43

Andrew Kor

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Samuel Newport · Answer 11 · 2017-12-07T15:15:07.990

0

Another interesting way to do it which would also allow more than just the last number to be taken would be:

int number = 124454;
int overflow = (int)Math.floor(number/(1*10^n))*10^n;

int firstDigits = number - overflow;
//Where n is the number of numbers you wish to conserve</code>

In the above example if n was 1 then the program would return: 4

If n was 3 then the program would return 454

edited Dec 07 '17 at 15:15

answered Dec 06 '17 at 01:37

Samuel Newport

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score -1 · Answer 12 · answered Jan 02 '17 at 20:14

Although the best way to do this is to use % if you insist on using strings this will work

public int lastDigit(int number)
{
return Integer.parseInt(String.valueOf(Integer.toString(number).charAt(Integer.toString(number).length() - 1)));
}

but I just wrote this for completeness. Do not use this code. it is just awful.

Gets last digit of a number

12 Answers12

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