I have numbers like 1100, 1002, 1022 etc. I would like to have the individual digits, for example for the first number 1100 I want to have 1, 1, 0, 0.
How can I get it in Java?
To do this, you will use the % (mod) operator.
int number; // = some int
while (number > 0) {
print( number % 10);
number = number / 10;
}
The mod operator will give you the remainder of doing int division on a number.
So,
10012 % 10 = 2
Because:
10012 / 10 = 1001, remainder 2
Note: As Paul noted, this will give you the numbers in reverse order. You will need to push them onto a stack and pop them off in reverse order.
Code to print the numbers in the correct order:
int number; // = and int
LinkedList<Integer> stack = new LinkedList<Integer>();
while (number > 0) {
stack.push( number % 10 );
number = number / 10;
}
while (!stack.isEmpty()) {
print(stack.pop());
}
Convert it to String and use String#toCharArray() or String#split().
String number = String.valueOf(someInt);
char[] digits1 = number.toCharArray();
// or:
String[] digits2 = number.split("(?<=.)");
In case you're already on Java 8 and you happen to want to do some aggregate operations on it afterwards, consider using String#chars() to get an IntStream out of it.
IntStream chars = number.chars();
How about this?
public static void printDigits(int num) {
if(num / 10 > 0) {
printDigits(num / 10);
}
System.out.printf("%d ", num % 10);
}
or instead of printing to the console, we can collect it in an array of integers and then print the array:
public static void main(String[] args) {
Integer[] digits = getDigits(12345);
System.out.println(Arrays.toString(digits));
}
public static Integer[] getDigits(int num) {
List<Integer> digits = new ArrayList<Integer>();
collectDigits(num, digits);
return digits.toArray(new Integer[]{});
}
private static void collectDigits(int num, List<Integer> digits) {
if(num / 10 > 0) {
collectDigits(num / 10, digits);
}
digits.add(num % 10);
}
If you would like to maintain the order of the digits from least significant (index[0]) to most significant (index[n]), the following updated getDigits() is what you need:
/**
* split an integer into its individual digits
* NOTE: digits order is maintained - i.e. Least significant digit is at index[0]
* #param num positive integer
* #return array of digits
*/
public static Integer[] getDigits(int num) {
if (num < 0) { return new Integer[0]; }
List<Integer> digits = new ArrayList<Integer>();
collectDigits(num, digits);
Collections.reverse(digits);
return digits.toArray(new Integer[]{});
}
I haven't seen anybody use this method, but it worked for me and is short and sweet:
int num = 5542;
String number = String.valueOf(num);
for(int i = 0; i < number.length(); i++) {
int j = Character.digit(number.charAt(i), 10);
System.out.println("digit: " + j);
}
This will output:
digit: 5
digit: 5
digit: 4
digit: 2
I noticed that there are few example of using Java 8 stream to solve your problem but I think that this is the simplest one:
int[] intTab = String.valueOf(number).chars().map(Character::getNumericValue).toArray();
To be clear:
You use String.valueOf(number) to convert int to String, then chars() method to get an IntStream (each char from your string is now an Ascii number), then you need to run map() method to get a numeric values of the Ascii number. At the end you use toArray() method to change your stream into an int[] array.
I see all the answer are ugly and not very clean.
I suggest you use a little bit of recursion to solve your problem. This post is very old, but it might be helpful to future coders.
public static void recursion(int number) {
if(number > 0) {
recursion(number/10);
System.out.printf("%d ", (number%10));
}
}
Output:
Input: 12345
Output: 1 2 3 4 5
simple solution
public static void main(String[] args) {
int v = 12345;
while (v > 0){
System.out.println(v % 10);
v /= 10;
}
}
// could be any num this is a randomly generated one
int num = (int) (Math.random() * 1000);
// this will return each number to a int variable
int num1 = num % 10;
int num2 = num / 10 % 10;
int num3 = num /100 % 10;
// you could continue this pattern for 4,5,6 digit numbers
// dont need to print you could then use the new int values man other ways
System.out.print(num1);
System.out.print("\n" + num2);
System.out.print("\n" + num3);
Since I don't see a method on this question which uses Java 8, I'll throw this in. Assuming that you're starting with a String and want to get a List<Integer>, then you can stream the elements like so.
List<Integer> digits = digitsInString.chars()
.map(Character::getNumericValue)
.boxed()
.collect(Collectors.toList());
This gets the characters in the String as a IntStream, maps those integer representations of characters to a numeric value, boxes them, and then collects them into a list.
Java 9 introduced a new Stream.iterate method which can be used to generate a stream and stop at a certain condition. This can be used to get all the digits in the number, using the modulo approach.
int[] a = IntStream.iterate(123400, i -> i > 0, i -> i / 10).map(i -> i % 10).toArray();
Note that this will get the digits in reverse order, but that can be solved either by looping through the array backwards (sadly reversing an array is not that simple), or by creating another stream:
int[] b = IntStream.iterate(a.length - 1, i -> i >= 0, i -> i - 1).map(i -> a[i]).toArray();
or
int[] b = IntStream.rangeClosed(1, a.length).map(i -> a[a.length - i]).toArray();
As an example, this code:
int[] a = IntStream.iterate(123400, i -> i > 0, i -> i / 10).map(i -> i % 10).toArray();
int[] b = IntStream.iterate(a.length - 1, i -> i >= 0, i -> i - 1).map(i -> a[i]).toArray();
System.out.println(Arrays.toString(a));
System.out.println(Arrays.toString(b));
Will print:
[0, 0, 4, 3, 2, 1]
[1, 2, 3, 4, 0, 0]
Easier way I think is to convert the number to string and use substring to extract and then convert to integer.
Something like this:
int digits1 =Integer.parseInt( String.valueOf(201432014).substring(0,4));
System.out.println("digits are: "+digits1);
ouput is
2014
I wrote a program that demonstrates how to separate the digits of an integer using a more simple and understandable approach that does not involve arrays, recursions, and all that fancy schmancy. Here is my code:
int year = sc.nextInt(), temp = year, count = 0;
while (temp>0)
{
count++;
temp = temp / 10;
}
double num = Math.pow(10, count-1);
int i = (int)num;
for (;i>0;i/=10)
{
System.out.println(year/i%10);
}
Suppose your input is the integer 123, the resulting output will be as follows:
1
2
3
Here is my answer, I did it for myself and I hope it's simple enough for those who don't want to use the String approach or need a more math-y solution:
public static void reverseNumber2(int number) {
int residual=0;
residual=number%10;
System.out.println(residual);
while (residual!=number) {
number=(number-residual)/10;
residual=number%10;
System.out.println(residual);
}
}
So I just get the units, print them out, substract them from the number, then divide that number by 10 - which is always without any floating stuff, since units are gone, repeat.
Java 8 solution to get digits as int[] from an integer that you have as a String:
int[] digits = intAsString.chars().map(i -> i - '0').toArray();
neither chars() nor codePoints() — the other lambda
String number = Integer.toString( 1100 );
IntStream.range( 0, number.length() ).map( i -> Character.digit( number.codePointAt( i ), 10 ) ).toArray(); // [1, 1, 0, 0]
Why don't you do:
String number = String.valueOf(input);
char[] digits = number.toCharArray();
Try this one.
const check = (num) => {
let temp = num
let result = []
while(temp > 0){
let a = temp%10;
result.push(a);
temp = (temp-a)/10;
}
return result;
}
check(98) //[ 8, 9 ]
public int[] getDigitsOfANumber(int number) {
String numStr = String.valueOf(number);
int retArr[] = new int[numStr.length()];
for (int i = 0; i < numStr.length(); i++) {
char c = numStr.charAt(i);
int digit = c;
int zero = (char) '0';
retArr[i] = digit - zero;
}
return retArr;
}
Integer.toString(1100) gives you the integer as a string. Integer.toString(1100).getBytes() to get an array of bytes of the individual digits.
Edit:
You can convert the character digits into numeric digits, thus:
String string = Integer.toString(1234);
int[] digits = new int[string.length()];
for(int i = 0; i<string.length(); ++i){
digits[i] = Integer.parseInt(string.substring(i, i+1));
}
System.out.println("digits:" + Arrays.toString(digits));
This uses the modulo 10 method to figure out each digit in a number greater than 0, then this will reverse the order of the array. This is assuming you are not using "0" as a starting digit.
This is modified to take in user input.
This array is originally inserted backwards, so I had to use the Collections.reverse() call to put it back into the user's order.
Scanner scanNumber = new Scanner(System.in);
int userNum = scanNumber.nextInt(); // user's number
// divides each digit into its own element within an array
List<Integer> checkUserNum = new ArrayList<Integer>();
while(userNum > 0) {
checkUserNum.add(userNum % 10);
userNum /= 10;
}
Collections.reverse(checkUserNum); // reverses the order of the array
System.out.print(checkUserNum);
Just to build on the subject, here's how to confirm that the number is a palindromic integer in Java:
public static boolean isPalindrome(int input) {
List<Integer> intArr = new ArrayList();
int procInt = input;
int i = 0;
while(procInt > 0) {
intArr.add(procInt%10);
procInt = procInt/10;
i++;
}
int y = 0;
int tmp = 0;
int count = 0;
for(int j:intArr) {
if(j == 0 && count == 0) {
break;
}
tmp = j + (tmp*10);
count++;
}
if(input != tmp)
return false;
return true;
}
I'm sure I can simplify this algo further. Yet, this is where I am. And it has worked under all of my test cases.
I hope this helps someone.
int number = 12344444; // or it Could be any valid number
int temp = 0;
int divider = 1;
for(int i =1; i< String.valueOf(number).length();i++)
{
divider = divider * 10;
}
while (divider >0) {
temp = number / divider;
number = number % divider;
System.out.print(temp +" ");
divider = divider/10;
}
Try this:
int num= 4321
int first = num % 10;
int second = ( num - first ) % 100 / 10;
int third = ( num - first - second ) % 1000 / 100;
int fourth = ( num - first - second - third ) % 10000 / 1000;
You will get first = 1, second = 2, third = 3 and fourth = 4 ....
Something like this will return the char[]:
public static char[] getTheDigits(int value){
String str = "";
int number = value;
int digit = 0;
while(number>0){
digit = number%10;
str = str + digit;
System.out.println("Digit:" + digit);
number = number/10;
}
return str.toCharArray();
}
As a noob, my answer would be:
String number = String.valueOf(ScannerObjectName.nextInt());
int[] digits = new int[number.length()];
for (int i = 0 ; i < number.length() ; i++)
int[i] = Integer.parseInt(digits.substring(i,i+1))
Now all the digits are contained in the "digits" array.
if digit is meant to be a Character
String numstr = Integer.toString( 123 );
Pattern.compile( "" ).splitAsStream( numstr ).map(
s -> s.charAt( 0 ) ).toArray( Character[]::new ); // [1, 2, 3]
and the following works correctly
numstr = "000123" gets [0, 0, 0, 1, 2, 3]
numstr = "-123" gets [-, 1, 2, 3]
A .NET solution using LINQ.
List<int> numbers = number.ToString().Select(x => x - 48).ToList();
I think this will be the most useful way to get digits:
public int[] getDigitsOf(int num)
{
int digitCount = Integer.toString(num).length();
if (num < 0)
digitCount--;
int[] result = new int[digitCount];
while (digitCount-- >0) {
result[digitCount] = num % 10;
num /= 10;
}
return result;
}
Then you can get digits in a simple way:
int number = 12345;
int[] digits = getDigitsOf(number);
for (int i = 0; i < digits.length; i++) {
System.out.println(digits[i]);
}
or more simply:
int number = 12345;
for (int i = 0; i < getDigitsOf(number).length; i++) {
System.out.println( getDigitsOf(number)[i] );
}
Notice the last method calls getDigitsOf method too much time. So it will be slower. You should create an int array and then call the getDigitsOf method once, just like in second code block.
In the following code, you can reverse to process. This code puts all digits together to make the number:
public int digitsToInt(int[] digits)
{
int digitCount = digits.length;
int result = 0;
for (int i = 0; i < digitCount; i++) {
result = result * 10;
result += digits[i];
}
return result;
}
Both methods I have provided works for negative numbers too.
see bellow my proposal with comments
int size=i.toString().length(); // the length of the integer (i) we need to split;
ArrayList<Integer> li = new ArrayList<Integer>(); // an ArrayList in whcih to store the resulting digits
Boolean b=true; // control variable for the loop in which we will reatrive step by step the digits
String number="1"; // here we will add the leading zero depending on the size of i
int temp; // the resulting digit will be kept by this temp variable
for (int j=0; j<size; j++){
number=number.concat("0");
}
Integer multi = Integer.valueOf(number); // the variable used for dividing step by step the number we received
while(b){
multi=multi/10;
temp=i/(multi);
li.add(temp);
i=i%(multi);
if(i==0){
b=false;
}
}
for(Integer in: li){
System.out.print(in.intValue()+ " ");
}
import java.util.Scanner;
class Test
{
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
int num=sc.nextInt();
System.out.println("Enter a number (-1 to end):"+num);
int result=0;
int i=0;
while(true)
{
int n=num%10;
if(n==-1){
break;
}
i++;
System.out.println("Digit"+i+" = "+n);
result=result*10+n;
num=num/10;
if(num==0)
{
break;
}
}
}
}
I tried to create a code to take in a whole number in Java and output it in binary. The problem would seem that the binary is printing out backward. For instance, 6 should output as 011 but comes out as 110.
import java.util.Scanner;
public class LabProgram {
public static void main(String[] args) {
int userNum;
Scanner in =new Scanner(System. in );
userNum = in.nextInt();
binary(userNum);
System.out.print("\n");
}
private static void binary(int userNum) {
int remainder;
while (userNum <= 1) {
System.out.print(userNum);
return;
}
remainder = userNum % 2;
binary(userNum >> 1);
System.out.print(remainder);
}
}
I tried incorporating a push stack to push the remainder into a stack that I can pull later, but couldn't quite get it to land.
private static void reverse(int userNum) {
String backwards;
while (userNum >= 0) {
backwards.push(int userNum);
System.out.println(backwards);
return;
}
}
It is part of a class assignment which asks the following.
Write a program that takes in a positive integer as input, and outputs a string of 1's and 0's representing the integer in binary. For an integer x, the algorithm is:
As long as x is greater than 0
Output x % 2 (remainder is either 0 or 1)
x = x / 2
Note: The above algorithm outputs the 0's and 1's in reverse order.
Ex: If the input is:
6
the output is:
011
6 in binary is 110; the algorithm outputs the bits in reverse.
These are the tests the program applies and my results.
Input 6
Your output binary is:110
Expected output 011
Input 19
Your output 10011
Expected output 11001
Input 255
Your output 11111111
Expected output 11111111
Any help or guidance in this, I would be greatly appreciative of it.
Per the requirement and not taking into consideration of negative numbers
import java.util.Scanner;
public class LabProgram {
public static void main(String[] args) {
int userNum;
Scanner scnr = new Scanner(System.in);
userNum = scnr.nextInt();
while(userNum > 0){
System.out.print(userNum % 2);
userNum = userNum / 2;
}
System.out.print("\n");
}
}
First using predefined method then a custom one.
public class IntToBinary {
public static void main(String[] args) {
int decimalNumber = 10;
System.out.println(Integer.toBinaryString(decimalNumber));
System.out.println(convertBinary(10));
}
public static String convertBinary(int num) {
StringBuilder sb = new StringBuilder();
int binary[] = new int[40];
int index = 0;
while (num > 0) {
binary[index++] = num % 2;
num = num / 2;
}
for (int i = index - 1; i >= 0; i--) {
sb.append(binary[i]);
}
return sb.toString();
}
}
Your program appears to work fine for positive values. However it does not handle negative numbers which have their own unique binary representation known as two's complement. You could do something like the following to accommodate:
private static void binary(int userNum) {
int remainder;
// while (userNum <= 1) {
// System.out.print(userNum);
// return;
// }
if (userNum == 0) {
return;
}
// simply mask off the bit instead of dividing by two
remainder = userNum & 1;
// and shift right thru the sign bit
binary(userNum >>> 1);
System.out.print(remainder);
}
}
binary(-6));
prints
11111111111111111111111111111010
And the reason these printed out in proper order is because your routine is recursive. That is a natural behavior of printing the values stored in the stack from a recursive procedure.
import java.util.Scanner;
public class Reverse_BinaryNum {
public static void main(String[] args) {
/* Type your code here. */
Scanner scnr = new Scanner(System.in);
int inputNum;
System.out.println("Enter the Digit : ");
inputNum = scnr.nextInt();
System.out.println("The Reverse Binary for the given Digit is : ");
while (inputNum > 0) {
System.out.print(inputNum % 2);
inputNum = inputNum / 2;
}
scnr.close();
}
}
Assume, you input the number 546, then you should find the product of its digits, which is 546=120, then multiply the digits of 120 until and so on, continue until you get a one digit number.
Here's the code I wrote, but the loop doesn't work correctly and I've tried to fix it, but nothing changed. Could you please help me?
import java.util.*;
public class Product {
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
int a = keyboard.nextInt();
System.out.println("Please input the number");
numberOfProducts(a);
}
public static void numberOfProducts(int n){
int product = 1;
while(n>10) {
int current = n;
while (current != 0) {
product = product * (current % 10);
current = current / 10;
n = product;
System.out.println(product);
}
}
}
}
For a different take on the solution you can use a recursive lambda
import java.util.Scanner;
import java.util.function.IntFunction;
public class Product {
// this simply reduces the number to the product of its digits.
static IntFunction<Integer> prod =
(a) -> a < 10 ? a : a % 10 * Product.prod.apply(a / 10);
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
System.out.print("Please input the number");
int n = keyboard.nextInt();
// Now continue applying the lambda until the result is < 10.
while (n > 10) {
n = prod.apply(n);
}
System.out.println(n);
}
}
I think you are looking something like the code below:
import java.util.Scanner;
public class Main {
public static int numberOfProducts(String number) {
int product = 1;
do {
for (int i = 0; i < number.length(); ++i){
// This line converts the every digit of the number string to an integer.
product *= number.charAt(i) - '0';
}
// Remove this line, if you don't want to print the product of each iteration.
System.out.println(number);
// Update number with the new product.
// This line converts the int product to a new string.
number = "" + product;
} while (product > 10);
return product;
}
public static void main(String[] args) {
System.out.print("Please input the number: ");
Scanner keyboard = new Scanner(System.in);
int a = keyboard.nextInt();
// Treat number as a string for easier indexing.
String number = "" + a;
System.out.println(numberOfProducts(number));
}
}
When the above code runs, with 546 as input, it outputs:
Please input the number: 546
546
120
0
After looking through your code, your issue seems to be in this expression:
current % 10.
The modulo operation gives you the remainder of a division by ten.
In the case of your input 120, the result of that operation would be 0.
Following the rest of your application logic, your iteration variable will be set to zero, ending your loop immediately.
I will not give you copy-paste code to fix this problem, as it seems like a programming course assignment. I will however help you solve it.
My suggested fix is to change your approach to this problem and not try to solve this the mathematical way, but rather in a way that takes advantage of the Java programing language.
You could change your input from an Integer to a String.
In which case, you can use String.length() to ensure your requirement is fulfilled when exiting the loop.
In your loop, you split the String into substrings of length 1. Afterwards, you just multiply these.
When the loop exits (because String length is no longer greater than 1) you will have your intended result.
Good luck!
Actually your code is very close to being correct, the only thing you're doing wrong is that you are not resetting the product variable between iterations. You simply need to move the int product = 1; instruction inside the outer while loop.
Also, if you want a single digit at the end, your condition should be while(n >= 10) or while(n > 9), since 10 is still 2 digits, so we need one more iteration !
Final remark: sometimes it's easier to break your code into smaller pieces. It is easier to understand and easier to test/debug ! For example you could have first created a function productOfDigits(n) that returns the result of a single iteration, and then another function productOfDigitsUntilSingleDigit(n) that repeatedly calls the previous function.
import java.util.*;
public class Product {
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
int a = keyboard.nextInt();
System.out.println("Please input the number");
numberOfProducts(a);
}
public static void numberOfProducts(int n){
int product = 1;
while (n != 0) {
product = product * (n % 10);
n = n / 10;
}
if(product > 10) {
System.out.println("Intermediate result="+product);
numberOfProducts(product);
}
else
System.out.println("Final 1 digit product="+product);
}
}
function getProductUntilSingleDigit(n) {
let product = 1;
while (n > 0 || product > 9) {
if (n == 0) {
n = product;
product = 1;
}
product = product * (n % 10);
n = Math.floor(n / 10);
}
return product;
}
console.log(getProductUntilSingleDigit(546));
I'm new in java.
I have to write a program that modifies an existing catalog (3-5 digits) number to a new catalog number by adding a number to the left of an existing catalog number according to these conditions:
The new number will be the largest number between the leftmost digit to the rightmost digit.
If leftmost digit equal to the rightmost, the new number will be 9.
The input should be 10 numbers and then to add a new number.
The problem is that, now, the method "newKatalogNumber" get the old catalog number, and return the left digit, i like the method return the new catalog code. for instance, if the method get 1234, she will return 41234, and it will be printed at the end of the main. I have not found a way to do it.
Is anybody have the clue how to do that?
I will be grateful.
This is my code:
import java.util.Scanner; // This program gets an old catalog number between 3 to 5 digits and change it to a new catalog number
// by adding a new digit to the left of the number
public class Number
{
public static int newKatalogNumber(int num)
{
while (num>=10)
{
num /= 10;
}
return num;
}
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
for (int oldNum=0; oldNum<10; oldNum++)
{
System.out.print("Insert old catalog Number: ");//The user insert the old catalog number
int catalogNum = input.nextInt();
int mostRight = catalogNum % 10;
int mostLeft = newKatalogNumber(catalogNum);
//Adding the new digit according to condition below:
if (mostRight>mostLeft)
{
System.out.println("Old catalog number is:"+catalogNum);
System.out.println("New catalog number is"+mostRight+catalogNum);
}
else if (mostLeft>mostRight)
{
System.out.println("Old catalog number is:"+catalogNum);
System.out.println("New catalog number is"+mostLeft+catalogNum);
}
else
{
System.out.println("Old catalog number is:"+catalogNum);
System.out.println("New catalog number is"+9+catalogNum);
}
}
}
}
If you have a number like 1234and you want to add at the beginning, let's say, a 5 then you should do:
1234 + 5 * 10000
And if you want to add the 5 at the end you should do:
1234 * 10 + 5
Notice in the first case the number of zeroes in 10000 equals the number of digits in the original number.
try this
public static int NewKatatlogNumber (int num)
{
int noOfDigitsInTheNumber = num + "".length();
int[] digitsArray = new int[noOfDigitsInTheNumber];
int index = digitsArray.length - 1;
//Extract digit by digit and populate the digitarray from right to left.
//If your num is 123 your array will [1,2,3]
while (num > 0)
{
int remainder = num % 10;
num = num / 10;
digitsArray[index--] = remainder;
}
//Find the number to add from the digit array.Apply your logic here
int numberToAddToTheLeft = getNumberToAdd(digitsArray);
int newNum = numberToAddToTheLeft;
//Construct the final token by prepending the digit you identified
for (int i = 0; i < digitsArray.length; i++)
{
int j = digitsArray[i];
newNum = newNum * 10 + i;
}
return newNum;
}
Convert the number to a string and then add "n" to the front of it. Here is an example using DrJava's interactions tab:
> Integer.parseInt("5" + Integer.toString(500))
5500
> Integer.parseInt(Integer.toString(5) + Integer.toString(500))
5500
But there is a much more succinct (but less efficient) way since adding "" to an int will convert it to a string:
> Integer.parseInt(5 + "" + 500)
5500
If you need to handle negative numbers you can do:
if (n>0)
System.out.println(Integer.parseInt(p + "" + n));
else
System.out.println(Integer.parseInt("-" + p + "" + -n));