I am still somewhat of a beginner to Java, but I need help with my code. I wanted to write an Armstrong Number checker.
An Armstrong number is one whose sum of digits raised to the power three equals the number itself. 371, for example, is an Armstrong number because 3^3 + 7^3 + 1^3 = 371.
If I understand this concept correctly, then my code should work fine, but I don't know where I made mistakes. I would appreciate if you could help correct my mistakes, but still kind of stick with my solution to the problem, unless my try is completely wrong or most of it needs to change.
Here is the code:
public class ArmstrongChecker {
boolean confirm = false;
Integer input;
String converter;
int indices;
int result = 1;
void ArmstrongCheck(Integer input) {
this.input = input;
converter = input.toString();
char[] array = converter.toCharArray();
indices = array.length;
result = (int) Math.pow(array[0], indices);
for (int i = 1; i < array.length; i++) {
result = result + (int) Math.pow(array[i], indices);
}
if (result == input) {
confirm = true;
System.out.println(confirm);
} else {
System.out.println(confirm);
}
}
}
For my tries I used '153' as an input. Thank you for your help!
You aren't summing the digits, but the numeric values of the characters representing them. You can convert such a character to its numeric value by subtracting the character '0':
int result = 0;
for(int i = 0; i < array.length; i++) {
result = result + (int) Math.pow(array[i] - '0', indices);
}
Having said that, it's arguably (probably?) more elegant to read the input as an actual int number and iterate its digits by taking the reminder of 10 on each iteration. The number of digits itself can be calculated using a base-10 log.
int temp = input;
int result = 0;
int indices = (int) Math.log10(input) + 1;
while (temp != 0) {
int digit = temp % 10;
result += (int) Math.pow(digit, indices);
temp /= 10;
}
There is a small logical mistake in your code, You're not converting the character to an integer instead you're doing something like
Math.pow('1', 3) -> Math.pow(49, 3) // what you're doing
Math.pow(1, 3) // what should be done
You should first convert the character to the string using any method below
result = (int) Math.pow(array[0],indices);
for(int i = 1;i<array.length;i++) {
result = result + (int) Math.pow(array[i],indices);
}
For converting char to integer
int x = Character.getNumericValue(array[i]);
or
int x = Integer.parseInt(String.valueOf(array[i]));
or
int x = array[i] - '0';
Alternatively
You can also check for Armstrong's number without any conversion, using the logic below
public class Armstrong {
public static void main(String[] args) {
int number = 153, num, rem, res = 0;
num = number;
while (num != 0)
{
rem = num % 10;
res += Math.pow(rem, 3);
num /= 10;
}
if(res == num)
System.out.println("YES");
else
System.out.println("NO");
}
}
For any int >= 0 you can do it like this.
Print all the Armstrong numbers less than 10_000.
for (int i = 0; i < 10_000; i++) {
if (isArmstrong(i)) {
System.out.println(i);
}
}
prints
0
1
2
3
4
5
6
7
8
9
153
370
371
407
1634
8208
9474
The key is to use Math.log10 to compute the number of digits in the candidate number. This must be amended by adding 1. So Math.log10(923) returns 2.965201701025912. Casting to an int and adding 1 would be 3 digits.
The number of digits is then the power used for computation.
Then it's just a matter of summing up the digits raised to that power. The method short circuits and returns false if the sum exceeds the number before all the digits are processed.
public static boolean isArmstrong(int v) {
if (v < 0) {
throw new IllegalArgumentException("Argument must >= 0");
}
int temp = v;
int power = (int)Math.log10(temp)+1;
int sum = 0;
while (temp > 0) {
sum += Math.pow(temp % 10, power);
if (sum > v) {
return false;
}
temp/= 10;
}
return v == sum;
}
I want to reverse an integer. I have trying by the following code, the problems occurs when the number is so large that its reverse gets larger than the int limit (for example 1534236469), so in that case it returns some garbage integer.
How can I make use of Integer.MIN_VALUE and Integer.MAX_VALUE to check if the reversed number is within the limits?
Note: it must use int variable type only.
int num = 1534236469;
int reverseInt = 0;
int multiplier = 1;
if ( num < 0 ) {
multiplier *= -1;
}
while ( num != 0 ) {
//get the last digit
int digit = num % 10;
//multiply the reverseInt by 10 and then add the last digit
reverseInt = (reverseInt * multiplier) + digit;
multiplier = 10;
num /= 10;
}
//how to fix this.
if (reverseInt < Integer.MIN_VALUE || reverseInt > Integer.MAX_VALUE ) {
System.out.println("Invalid number");
} else {
System.out.println("Reversed integer is " + reverseInt);
}
You just have to make the check before you add the next digit, so you can simply move your check up to the actual calculation. If your current reversed positive number is greater than Integer.MAX_VALUE/10, then you can't add another digit. Likewise with negative numbers.
The only thing I did was move this part of your code up:
if (reverseInt < Integer.MIN_VALUE || reverseInt > Integer.MAX_VALUE ) {
System.out.println("Invalid number");
} else {
then I added the /10 and a return statement, since the program must end when there's an overflow:
public class StackOverflowTest {
public static void main(String[] args) {
int num = -1534236469;
int reverseInt = 0;
int multiplier = 1;
if ( num < 0 ) {
multiplier *= -1;
}
while ( num != 0 ) {
// if this next step will push is into overflow, then stop:
if (reverseInt < Integer.MIN_VALUE/10 || reverseInt > Integer.MAX_VALUE/10) {
System.out.println("Invalid number");
return;
} else {
//get the last digit
int digit = num % 10;
// multiply the reverseInt by 10 and then add the last digit
reverseInt = (reverseInt * multiplier) + digit;
}
multiplier = 10;
num /= 10;
}
System.out.println("Reversed integer is " + reverseInt);
}
}
Alternatively you can treat it as a String, and then simply reverse the String (along with fiddling with the sign):
public class StackOverflowTest {
public static void main(String[] args) {
reverse(1534236469);
System.out.println();
reverse(-153423646);
}
public static void reverse(int num) {
System.out.println("int = " + num);
int number = num < 0 ? -num : num; // remove the sign
String reverse = new StringBuilder(String.valueOf(number)).reverse().toString();
System.out.println("reverse String: " + reverse);
try {
int reversed = Integer.parseInt(reverse);
reversed = num < 0 ? -reversed : reversed; // get back the sign
System.out.println("Reversed integer is " + reversed);
} catch (NumberFormatException ex) {
System.out.println("Invalid number");
}
}
}
Prints:
int = 1534236469
reverse String: 9646324351
Invalid number
int = -153423646
reverse String: 646324351
Reversed integer is -646324351
In Java 8+, change to:
reverseInt = Math.addExact(Math.multiplyExact(reverseInt, multiplier), digit);
The code will now throw ArithmeticException if the result overflows.
Whenever the value of Integer.MAX_VALUE is exceeded, it starts from Integer.MIN_VALUE i.e. Integer.MAX_VALUE + 1 evaluates to Integer.MIN_VALUE. You can use this feature to get the solution you are looking for.
Simply multiply reverseInt in a loop from 1 to 10 and if the value becomes negative, it means that the next value (reverseInt * multiplier) will exceed Integer.MAX_VALUE. Also, you need to add reverseInt * multiplier in a loop to a value from 0 to 9 and if the value becomes negative, it means that the next value (reverseInt * multiplier + digit) will exceed Integer.MAX_VALUE.
public class Main {
public static void main(String[] args) throws InterruptedException {
int num = 1534236469;
boolean valid = true;
int reverseInt = 0;
int multiplier = 1;
if (num < 0) {
multiplier *= -1;
}
while (num != 0) {
// get the last digit
int digit = num % 10;
for (int i = 1; i <= multiplier; i++) {
for (int j = 0; j <= 9; j++) {
if (reverseInt * i < 0 || (reverseInt * i + j) < 0) {
System.out.println("Invalid number");
valid = false;
break;
}
}
if (!valid) {
break;
}
}
if (!valid) {
break;
}
// multiply the reverseInt by 10 and then add the last digit
reverseInt = reverseInt * multiplier + digit;
multiplier = 10;
num /= 10;
}
if (valid) {
System.out.println("Reversed integer is " + reverseInt);
}
}
}
Output:
Invalid number
Below is my code.
package com.ofss.java.examples;
import java.util.Scanner;
class ArmstrongNumber {
public static void main(String[] args) {
int c = 0, a;
int n1, n2;//Range in which armstrong number need to find
Scanner s = new Scanner(System.in);
System.out.println("Enter the first number");
n1 = s.nextInt();
System.out.println("Enter the second number");
n2 = s.nextInt();
for (int i = n1; i <= n2; ++i) {
while (i > 0) {
a = i % 10;
System.out.println(a);
i = i / 10;
System.out.println(i);
c = c + (a * a * a);
System.out.println(c);
}
if (i == c)
System.out.println(c + "armstrong number");
else
System.out.println(c + "Not armstrong number");
}
}
}
I am getting incorrect results after executing. Code runs for infinite number till you stops it. It must print number between 151-154 (153 as armstrong).
Also, it is incorrectly printing 153 as not Armstrong number.
Armstrong Number
...is a number that is the sum of its own digits each raised to the power of the number of digits.
You should not change i since this is also used in
for (int i = n1; i <= n2; ++i)
Or you will probably never exit that loop eiter since you expect i to be negative at the end of the first iteration. Hard to increment until it reach n2.
Use a different variable to keep track of i safely.
int j = i;
while(j > 0) ...
About Armstrong number:
Armstrong number is a number that is the sum of its own digits each raised to the power of the number of digits
You need to put each digit to the power of the length of the number (the number of digit).
153 = 1^3 + 5^3 + 3^3
1634 = 1^4 + 6^4 + 3^4 + 4^4
Here is the method for it :
public static boolean isArmstrongNumber(int number){
int power = Integer.toString(number).length(); //just to get the number of digit...
int tmp = number;
int digit , sum = 0;
while(tmp > 0){
digit = tmp % 10;
sum += Math.round(Math.pow(digit , power));
tmp /= 10;
}
return sum == number;
}
Using this check from 0 to 10.000 gives :
0
1
2
3
4
5
6
7
8
9
153
370
371
407
1634
8208
9474
Same as Wikipedia :
The sequence of base 10 narcissistic numbers starts: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407, 1634, 8208, 9474, ...
Note that using a method remove the risk of forgetting to reset your variable like c in your case. Correcting this would give you a few more "correct" results (well the one with 3 digits)
You can also use less mathematics to read the number and use char[], remember that you need to substract '0' value to get the numeric value for a character :
public static boolean isArmstrongNumber(int number){
char[] digits = Integer.toString(number).toCharArray();
int power = digits.length;
int sum = 0;
for(char c : digits){
int digit = c - '0';
sum += Math.round(Math.pow(digit, power));
}
return sum == number;
}
There are two things.
you are updating i everytime as you have used it in while, so use different variable than i for this calculation.
int num = i;
c is used to compare sum of cube is same as number but you are not resetting it once the one iteration is over. so make c=0 inside for loop.
c = 0;
Also while printing you are using c, there should be i which is correct and real number.
Below is the working code you may try.
public static void main(String[] args)
{
int c = 0, a;
int n1, n2;//Range in which armstrong number need to find
Scanner s = new Scanner(System.in);
System.out.println("Enter the first number");
n1 = s.nextInt();
System.out.println("Enter the second number");
n2 = s.nextInt();
for (int i = n1; i <= n2; ++i)
{
int num = i;
while (num > 0)
{
a = num % 10;
num = num / 10;
c = c + (a * a * a);
}
if (i == c)
System.out.println(i + "armstrong number");
else
System.out.println(i + "Not armstrong number");
c = 0;
}
}
public class ArmstrongNumber {
private final int n1, n2;
public ArmstrongNumber(int n1, int n2) {
this.n1 = n1;
this.n2 = n2;
}
protected static boolean isArmstrong(int n) {
if(n < 0)
return false;
int remaining=n;
int sumCube=0;
while (remaining>0) {
int d = remaining % 10;
sumCube += cube(d);
remaining /= 10;
}
return n == sumCube;
}
private static int cube(int d) {
return d*d*d;
}
public Integer[] find() {
List<Integer> results = new ArrayList<>();
for (int i = n1; i <= n2; ++i)
{
if (isArmstrong(i))
results.add(i);
}
return results.toArray(new Integer[0]);
}
}
There are lot of things to improve in your code. One of the working logics as below:
for (int i = n1; i <= n2; ++i) {
int sum = 0, remainder = 0, digits = 0, temp = 0;
temp = i;
while (temp != 0) {
digits++;
temp = temp / 10;
}
temp = i;
while (temp != 0) {
remainder = temp % 10;
sum = sum + (int) Math.pow(remainder, digits);
temp = temp / 10;
}
if (i == sum)
System.out.println(i + " is an Armstrong number.");
else
System.out.println(i + " isn't an Armstrong number.");
}
I am trying to create a program to validate 10 to 12 digit long number sequences based on the luhn algorithm, but my program keeps on telling me that every number is invalid even though they're not.
This number should be valid, but my code doesn't think so: 8112189876
This number should not be valid, which my program agrees with, as it thinks every number is invalid: 8112189875
Here is my code:
static void luhn(){
System.out.print("Enter number to validate:\n");
String pnr = input.nextLine();
int length = pnr.length();
int sum = 0;
for (int i = 1, pos = length - 1; i < 10; i++, pos--){
char tmp = pnr.charAt(pos);
int num = tmp - 0
int product;
if (i % 2 != 0){
product = num * 1;
}
else{
product = num * 2;
}
if (product > 9)
product -= 9;
sum+= product;
boolean valid = (sum % 10 == 0);
if (valid){
System.out.print("Valid!\r");
}
else{
System.out.print("Invalid!");
}
}
}
use org.apache.commons.validator.routines.checkdigit.LuhnCheckDigit.LUHN_CHECK_DIGIT.isValid(number)
Maven Dependency:
<dependency>
<groupId>commons-validator</groupId>
<artifactId>commons-validator</artifactId>
<version>1.5.1</version>
</dependency>
The first thing I see is that you have:
int num = tmp - 0
You should instead have:
int num = tmp - '0';
Secondly, you should be validating your sum outside of the for loop, because you only care about the sum after processing all the digits.
Thirdly, you are starting from the end of the number, and you are not including the first number of your string. Why not use i for both tasks?
Resulting (working) method:
static void luhn(){
System.out.print("Enter number to validate:\n");
String pnr = input.nextLine();
// this only works if you are certain all input will be at least 10 characters
int extraChars = pnr.length() - 10;
if (extraChars < 0) {
throw new IllegalArgumentException("Number length must be at least 10 characters!");
}
pnr = pnr.substring(extraChars, 10 + extraChars);
int sum = 0;
// #3: removed pos
for (int i = 0; i < pnr.length(); i++){
char tmp = pnr.charAt(i);
// #1: fixed the '0' problem
int num = tmp - '0';
int product;
if (i % 2 != 0){
product = num * 1;
}
else{
product = num * 2;
}
if (product > 9)
product -= 9;
sum+= product;
}
// #2: moved check outside for loop
boolean valid = (sum % 10 == 0);
if (valid){
System.out.print("Valid!\r");
}
else{
System.out.print("Invalid!");
}
}
Stylistically, this method would be more useful if, instead of method signature
static void luhn() {
it instead had method signature
static boolean luhn(String input) {
This easily allows your code to get the String from ANY source (a file, hardcoded, etc.) and do anything with the result (print a message as yours does, or do something else). Obviously you would move the System.out.print, input.nextLine(), and if(valid) bits of code outside of this method.
Full refactored program:
import java.util.Scanner;
public class Luhn {
private static Scanner input;
public static void main(String... args) {
input = new Scanner(System.in);
System.out.print("Enter number to validate:\n");
String pnr = input.nextLine();
boolean result = luhn(pnr);
printMessage(result);
input.close();
}
static boolean luhn(String pnr){
// this only works if you are certain all input will be at least 10 characters
int extraChars = pnr.length() - 10;
if (extraChars < 0) {
throw new IllegalArgumentException("Number length must be at least 10 characters!");
}
pnr = pnr.substring(extraChars, 10 + extraChars);
int sum = 0;
for (int i = 0; i < pnr.length(); i++){
char tmp = pnr.charAt(i);
int num = tmp - '0';
int product;
if (i % 2 != 0){
product = num * 1;
}
else{
product = num * 2;
}
if (product > 9)
product -= 9;
sum+= product;
}
return (sum % 10 == 0);
}
private static void printMessage(boolean valid) {
if (valid){
System.out.print("Valid!\r");
}
else{
System.out.print("Invalid!");
}
}
}
I use this function in an app for checking card number validity :
public static boolean Check(String ccNumber)
{
int sum = 0;
boolean alternate = false;
for (int i = ccNumber.length() - 1; i >= 0; i--)
{
int n = Integer.parseInt(ccNumber.substring(i, i + 1));
if (alternate)
{
n *= 2;
if (n > 9)
{
n = (n % 10) + 1;
}
}
sum += n;
alternate = !alternate;
}
return (sum % 10 == 0);
}
Hope it helps,
If you use Java 10 or higher, you can use the following code:
public static boolean luhn(String s) {
IntUnaryOperator sumDigits = n -> n / 10 + n % 10;
var digits = s.chars()
.map(Character::getNumericValue)
.toArray();
return IntStream.rangeClosed(1, digits.length)
.map(i -> digits.length - i)
.map(i -> i % 2 == 0 ? digits[i] : sumDigits.applyAsInt(digits[i] * 2))
.sum() % 10 == 0;
}
It's the functional approach to this algorithm.
You should be subtracting '0' from tmp, not 0. Subtracting 0 returns the ASCII value, which you don't want.
Here's some functions I wrote to both calculate the check digit for a given number and to verify a given number sequence and extract the number from it.
To calculate the check digit for a given number:
/**
* Generates the check digit for a number using Luhn's algorithm described in detail at the following link:
* https://en.wikipedia.org/wiki/Luhn_algorithm
*
* In short the digit is calculated like so:
* 1. From the rightmost digit moving left, double the value of every second digit. If that value is greater than 9,
* subtract 9 from it.
* 2. Sum all of the digits together
* 3. Multiply the sum by 9 and the check digit will be that value modulo 10.
*
* #param number the number to get the Luhn's check digit for
* #return the check digit for the given number
*/
public static int calculateLuhnsCheckDigit(final long number) {
int sum = 0;
boolean alternate = false;
String digits = Long.toString(number);
for (int i = digits.length() - 1; i >= 0; --i) {
int digit = Character.getNumericValue(digits.charAt(i)); // get the digit at the given index
digit = (alternate = !alternate) ? (digit * 2) : digit; // double every other digit
digit = (digit > 9) ? (digit - 9) : digit; // subtract 9 if the value is greater than 9
sum += digit; // add the digit to the sum
}
return (sum * 9) % 10;
}
To verify a sequence of digits using Luhn's algorithm and extract the number:
/**
* Verifies that a given number string is valid according to Luhn's algorithm, which is described in detail here:
* https://en.wikipedia.org/wiki/Luhn_algorithm
*
* In short, validity of the number is determined like so:
* 1. From the rightmost digit (the check digit) moving left, double the value of every second digit. The check
* digit is not doubled; the first digit doubled is the one immediately to the left of the check digit. If that
* value is greater than 9, subtract 9 from it.
* 2. Sum all of the digits together
* 3. If the sum modulo 10 is equal to 0, then the number is valid according to Luhn's algorithm
*
* #param luhnsNumber the number string to verify and extract the number from
* #return an empty Optional if the given string was not valid according to Luhn's algorithm
* an Optional containing the number verified by Luhn's algorithm if the given string passed the check
*/
public static Optional<Long> extractLuhnsNumber(final String luhnsNumber) {
int sum = 0;
boolean alternate = true;
Long number = Long.parseLong(luhnsNumber.substring(0, luhnsNumber.length() - 1));
for (int i = luhnsNumber.length() - 1; i >= 0; --i) {
int digit = Character.getNumericValue(luhnsNumber.charAt(i)); // get the digit at the given index
digit = (alternate = !alternate) ? (digit * 2) : digit; // double every other digit
digit = (digit > 9) ? (digit - 9) : digit; // subtract 9 if the value is greater than 9
sum += digit; // add the digit to the sum
}
return (sum % 10 == 0) ? Optional.of(number) : Optional.empty();
}
Newcomers to this post/question can check appropriate Wikipedia page for solution. Below is the Java code copy-pasted from there.
public class Luhn
{
public static boolean check(String ccNumber)
{
int sum = 0;
boolean alternate = false;
for (int i = ccNumber.length() - 1; i >= 0; i--)
{
int n = Integer.parseInt(ccNumber.substring(i, i + 1));
if (alternate)
{
n *= 2;
if (n > 9)
{
n = (n % 10) + 1;
}
}
sum += n;
alternate = !alternate;
}
return (sum % 10 == 0);
}
}
package randomNumGen;
public class JavaLuhnAlgorithm {
public static void main(String[] args) {
// TODO Auto-generated method stub
validateCreditCardNumber("8112189876");
String imei = "012850003580200";
validateCreditCardNumber(imei);
}
private static void validateCreditCardNumber(String str) {
int[] ints = new int[str.length()];
for (int i = 0; i < str.length(); i++) {
ints[i] = Integer.parseInt(str.substring(i, i + 1));
}
for (int i = ints.length - 2; i >= 0; i = i - 2) {
int j = ints[i];
j = j * 2;
if (j > 9) {
j = j % 10 + 1;
}
ints[i] = j;
}
int sum = 0;
for (int i = 0; i < ints.length; i++) {
sum += ints[i];
}
if (sum % 10 == 0) {
System.out.println(str + " is a valid credit card number");
} else {
System.out.println(str + " is an invalid credit card number");
}
}
}
Can anyone explain to me how to reverse an integer without using array or String. I got this code from online, but not really understand why + input % 10 and divide again.
while (input != 0) {
reversedNum = reversedNum * 10 + input % 10;
input = input / 10;
}
And how to do use this sample code to reverse only odd number. Example I got this input 12345, then it will reverse the odd number to output 531.
Java reverse an int value - Principles
Modding (%) the input int by 10 will extract off the rightmost digit. example: (1234 % 10) = 4
Multiplying an integer by 10 will "push it left" exposing a zero to the right of that number, example: (5 * 10) = 50
Dividing an integer by 10 will remove the rightmost digit. (75 / 10) = 7
Java reverse an int value - Pseudocode:
a. Extract off the rightmost digit of your input number. (1234 % 10) = 4
b. Take that digit (4) and add it into a new reversedNum.
c. Multiply reversedNum by 10 (4 * 10) = 40, this exposes a zero to the right of your (4).
d. Divide the input by 10, (removing the rightmost digit). (1234 / 10) = 123
e. Repeat at step a with 123
Java reverse an int value - Working code
public int reverseInt(int input) {
long reversedNum = 0;
long input_long = input;
while (input_long != 0) {
reversedNum = reversedNum * 10 + input_long % 10;
input_long = input_long / 10;
}
if (reversedNum > Integer.MAX_VALUE || reversedNum < Integer.MIN_VALUE) {
throw new IllegalArgumentException();
}
return (int) reversedNum;
}
You will never do anything like this in the real work-world. However, the process by which you use to solve it without help is what separates people who can solve problems from the ones who want to, but can't unless they are spoon fed by nice people on the blogoblags.
I am not clear about your Odd number.
The way this code works is (it is not a Java specific algorithm)
Eg.
input =2345
first time in the while loop
rev=5 input=234
second time
rev=5*10+4=54 input=23
third time
rev=54*10+3 input=2
fourth time
rev=543*10+2 input=0
So the reversed number is 5432.
If you just want only the odd numbers in the reversed number then.
The code is:
while (input != 0) {
last_digit = input % 10;
if (last_digit % 2 != 0) {
reversedNum = reversedNum * 10 + last_digit;
}
input = input / 10;
}
Simply you can use this
public int getReverseInt(int value) {
int resultNumber = 0;
for (int i = value; i !=0; i /= 10) {
resultNumber = resultNumber * 10 + i % 10;
}
return resultNumber;
}
You can use this method with the given value which you want revers.
while (num != 0) {
rev = rev * 10 + num % 10;
num /= 10;
}
That is the solution I used for this problem, and it works fine.
More details:
num % 10
This statement will get you the last digit from the original number.
num /= 10
This statement will eliminate the last digit from the original number, and hence we are sure that while loop will terminate.
rev = rev * 10 + num % 10
Here rev*10 will shift the value by left and then add the last digit from the original.
If the original number was 1258, and in the middle of the run time we have rev = 85, num = 12 so:
num%10 = 2
rev*10 = 850
rev*10 + num%10 = 852
int aa=456;
int rev=Integer.parseInt(new StringBuilder(aa+"").reverse());
import java.util.Scanner;
public class Reverse_order_integer {
private static Scanner scan;
public static void main(String[] args) {
System.out.println("\t\t\tEnter Number which you want to reverse.\n");
scan = new Scanner(System.in);
int number = scan.nextInt();
int rev_number = reverse(number);
System.out.println("\t\t\tYour reverse Number is = \"" + rev_number
+ "\".\n");
}
private static int reverse(int number) {
int backup = number;
int count = 0;
while (number != 0) {
number = number / 10;
count++;
}
number = backup;
int sum = 0;
for (int i = count; i > 0; i--) {
int sum10 = 1;
int last = number % 10;
for (int j = 1; j < i; j++) {
sum10 = sum10 * 10;
}
sum = sum + (last * sum10);
number = number / 10;
}
return sum;
}
}
See to get the last digit of any number we divide it by 10 so we either achieve zero or a digit which is placed on last and when we do this continuously we get the whole number as an integer reversed.
int number=8989,last_num,sum=0;
while(number>0){
last_num=number%10; // this will give 8989%10=9
number/=10; // now we have 9 in last and now num/ by 10= 898
sum=sum*10+last_number; // sum=0*10+9=9;
}
// last_num=9. number= 898. sum=9
// last_num=8. number =89. sum=9*10+8= 98
// last_num=9. number=8. sum=98*10+9=989
// last_num=8. number=0. sum=989*10+8=9898
// hence completed
System.out.println("Reverse is"+sum);
public static void main(String args[]) {
int n = 0, res = 0, n1 = 0, rev = 0;
int sum = 0;
Scanner scan = new Scanner(System.in);
System.out.println("Please Enter No.: ");
n1 = scan.nextInt(); // String s1=String.valueOf(n1);
int len = (n1 == 0) ? 1 : (int) Math.log10(n1) + 1;
while (n1 > 0) {
rev = res * ((int) Math.pow(10, len));
res = n1 % 10;
n1 = n1 / 10;
// sum+=res; //sum=sum+res;
sum += rev;
len--;
}
// System.out.println("sum No: " + sum);
System.out.println("sum No: " + (sum + res));
}
This will return reverse of integer
Just to add on, in the hope to make the solution more complete.
The logic by #sheki already gave the correct way of reversing an integer in Java. If you assume the input you use and the result you get always fall within the range [-2147483648, 2147483647], you should be safe to use the codes by #sheki. Otherwise, it'll be a good practice to catch the exception.
Java 8 introduced the methods addExact, subtractExact, multiplyExact and toIntExact. These methods will throw ArithmeticException upon overflow. Therefore, you can use the below implementation to implement a clean and a bit safer method to reverse an integer. Generally we can use the mentioned methods to do mathematical calculation and explicitly handle overflow issue, which is always recommended if there's a possibility of overflow in the actual usage.
public int reverse(int x) {
int result = 0;
while (x != 0){
try {
result = Math.multiplyExact(result, 10);
result = Math.addExact(result, x % 10);
x /= 10;
} catch (ArithmeticException e) {
result = 0; // Exception handling
break;
}
}
return result;
}
Java solution without the loop. Faster response.
int numberToReverse;//your number
StringBuilder sb=new StringBuilder();
sb.append(numberToReverse);
sb=sb.reverse();
String intermediateString=sb.toString();
int reversedNumber=Integer.parseInt(intermediateString);
import java.io.BufferedReader;
import java.io.InputStreamReader;
public class intreverse
{
public static void main(String...a)throws Exception
{
int no;
int rev = 0;
System.out.println("Enter The no to be reversed");
InputStreamReader str=new InputStreamReader(System.in);
BufferedReader br =new BufferedReader(str);
no=Integer.parseInt(br.readLine().toString());
while(no!=0)
{
rev=rev*10+no%10;
no=no/10;
}
System.out.println(rev);
}
}
public static int reverse(int x) {
boolean negetive = false;
if (x < 0) {
x = Math.abs(x);
negative = true;
}
int y = 0, i = 0;
while (x > 0) {
if (i > 0) {
y *= 10;
}
y += x % 10;
x = x / 10;
i++;
}
return negative ? -y : y;
}
Here is a complete solution(returns 0 if number is overflown):
public int reverse(int x) {
boolean flag = false;
// Helpful to check if int is within range of "int"
long num = x;
// if the number is negative then turn the flag on.
if(x < 0) {
flag = true;
num = 0 - num;
}
// used for the result.
long result = 0;
// continue dividing till number is greater than 0
while(num > 0) {
result = result*10 + num%10;
num= num/10;
}
if(flag) {
result = 0 - result;
}
if(result > Integer.MAX_VALUE || result < Integer.MIN_VALUE) {
return 0;
}
return (int) result;
}
Scanner input = new Scanner(System.in);
System.out.print("Enter number :");
int num = input.nextInt();
System.out.print("Reverse number :");
int value;
while( num > 0){
value = num % 10;
num /= 10;
System.out.print(value); //value = Reverse
}
int convert (int n)
{
long val = 0;
if(n==0)
return 0;
for(int i = 1; n > exponent(10, (i-1)); i++)
{
int mod = n%( (exponent(10, i))) ;
int index = mod / (exponent(10, i-1));
val *= 10;
val += index;
}
if (val < Integer.MIN_VALUE || val > Integer.MAX_VALUE)
{
throw new IllegalArgumentException
(val + " cannot be cast to int without changing its value.");
}
return (int) val;
}
static int exponent(int m, int n)
{
if(n < 0)
return 0;
if(0 == n)
return 1;
return (m * exponent(m, n-1));
}
It's good that you wrote out your original code. I have another way to code this concept of reversing an integer. I'm only going to allow up to 10 digits. However, I am going to make the assumption that the user will not enter a zero.
if((inputNum <= 999999999)&&(inputNum > 0 ))
{
System.out.print("Your number reversed is: ");
do
{
endInt = inputNum % 10; //to get the last digit of the number
inputNum /= 10;
system.out.print(endInt);
}
While(inputNum != 0);
System.out.println("");
}
else
System.out.println("You used an incorrect number of integers.\n");
System.out.println("Program end");
Even if negative integer is passed then it will give the negative integer
Try This...
public int reverse(int result) {
long newNum=0,old=result;
result=(result>0) ? result:(0-result);
while(result!=0){
newNum*=10;
newNum+=result%10;
result/=10;
if(newNum>Integer.MAX_VALUE||newNum<Integer.MIN_VALUE)
return 0;
}
if(old > 0)
return (int)newNum;
else if(old < 0)
return (int)(newNum*-1);
else
return 0;
}
This is the shortest code to reverse an integer
int i=5263;
System.out.println(Integer.parseInt(new StringBuffer(String.valueOf(i) ).reverse().toString()));
123 maps to 321, which can be calculated as 3*(10^2)+2*(10^1)+1
Two functions are used to calculate (10^N). The first function calculates the value of N. The second function calculates the value for ten to power N.
Function<Integer, Integer> powerN = x -> Double.valueOf(Math.log10(x)).intValue();
Function<Integer, Integer> ten2powerN = y -> Double.valueOf(Math.pow(10, y)).intValue();
// 123 => 321= 3*10^2 + 2*10 + 1
public int reverse(int number) {
if (number < 10) {
return number;
} else {
return (number % 10) * powerN.andThen(ten2powerN).apply(number) + reverse(number / 10);
}
}
If the idea is not to use arrays or string, reversing an integer has to be done by reading the digits of a number from the end one at a time. Below explanation is provided in detail to help the novice.
pseudocode :
lets start with reversed_number = 0 and some value for original_number which needs to be reversed.
the_last_digit = original_number % 10 (i.e, the reminder after dividing by 10)
original_number = original_number/10 (since we already have the last digit, remove the last digit from the original_number)
reversed_number = reversed_number * 10 + last_digit (multiply the reversed_number with 10, so as to add the last_digit to it)
repeat steps 2 to 4, till the original_number becomes 0. When original_number = 0, reversed_number would have the reverse of the original_number.
More info on step 4: If you are provided with a digit at a time, and asked to append it at the end of a number, how would you do it - by moving the original number one place to the left so as to accommodate the new digit. If number 23 has to become 234, you multiply 23 with 10 and then add 4.
234 = 23x10 + 4;
Code:
public static int reverseInt(int original_number) {
int reversed_number = 0;
while (original_number > 0) {
int last_digit = original_number % 10;
original_number = original_number / 10;
reversed_number = reversed_number * 10 + last_digit;
}
return reversed_number;
}
It is an outdated question, but as a reference for others
First of all reversedNum must be initialized to 0;
input%10 is used to get the last digit from input
input/10 is used to get rid of the last digit from input, which you have added to the reversedNum
Let's say input was 135
135 % 10 is 5
Since reversed number was initialized to 0
now reversedNum will be 5
Then we get rid of 5 by dividing 135 by 10
Now input will be just 13
Your code loops through these steps until all digits are added to the reversed number or in other words untill input becomes 0.
while (input != 0) {
reversedNum = reversedNum * 10 + input % 10;
input = input / 10;
}
let a number be 168,
+ input % 10 returns last digit as reminder i.e. 8 but next time it should return 6,hence number must be reduced to 16 from 168, as divide 168 by 10 that results to 16 instead of 16.8 as variable input is supposed to be integer type in the above program.
If you wanna reverse any number like 1234 and you want to revers this number to let it looks like 4321. First of all, initialize 3 variables int org ; int reverse = 0; and int reminder ;
then put your logic like
Scanner input = new Scanner (System.in);
System.out.println("Enter number to reverse ");
int org = input.nextInt();
int getReminder;
int r = 0;
int count = 0;
while (org !=0){
getReminder = org%10;
r = 10 * r + getReminder;
org = org/10;
}
System.out.println(r);
}
A method to get the greatest power of ten smaller or equal to an integer: (in recursion)
public static int powerOfTen(int n) {
if ( n < 10)
return 1;
else
return 10 * powerOfTen(n/10);
}
The method to reverse the actual integer:(in recursion)
public static int reverseInteger(int i) {
if (i / 10 < 1)
return i ;
else
return i%10*powerOfTen(i) + reverseInteger(i/10);
}
You can use recursion to solve this.
first get the length of an integer number by using following recursive function.
int Length(int num,int count){
if(num==0){
return count;
}
else{
count++;
return Lenght(num/10,count);
}
}
and then you can simply multiply remainder of a number by 10^(Length of integer - 1).
int ReturnReverse(int num,int Length,int reverse){
if(Length!=0){
reverse = reverse + ((num%10) * (int)(Math.pow(10,Length-1)));
return ReturnReverse(num/10,Length-1,reverse);
}
return reverse;
}
The whole Source Code :
import java.util.Scanner;
public class ReverseNumbers {
int Length(int num, int count) {
if (num == 0) {
return count;
} else {
return Length(num / 10, count + 1);
}
}
int ReturnReverse(int num, int Length, int reverse) {
if (Length != 0) {
reverse = reverse + ((num % 10) * (int) (Math.pow(10, Length - 1)));
return ReturnReverse(num / 10, Length - 1, reverse);
}
return reverse;
}
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int N = scanner.nextInt();
ReverseNumbers reverseNumbers = new ReverseNumbers();
reverseNumbers.ReturnReverse(N, reverseNumbers.Length(N, 0), reverseNumbers.ReturnReverse(N, reverseNumbers.Length(N, 0), 0));
scanner.close();
}
}
public int getReverseNumber(int number)
{
int reminder = 0, result = 0;
while (number !=0)
{
if (number >= 10 || number <= -10)
{
reminder = number % 10;
result = result + reminder;
result = result * 10;
number = number / 10;
}
else
{
result = result + number;
number /= 10;
}
}
return result;
}
// The above code will work for negative numbers also
Reversing integer
int n, reverse = 0;
Scanner in = new Scanner(System.in);
n = in.nextInt();
while(n != 0)
{
reverse = reverse * 10;
reverse = reverse + n%10;
n = n/10;
}
System.out.println("Reverse of the number is " + reverse);
public static int reverseInt(int i) {
int reservedInt = 0;
try{
String s = String.valueOf(i);
String reversed = reverseWithStringBuilder(s);
reservedInt = Integer.parseInt(reversed);
}catch (NumberFormatException e){
System.out.println("exception caught was " + e.getMessage());
}
return reservedInt;
}
public static String reverseWithStringBuilder(String str) {
System.out.println(str);
StringBuilder sb = new StringBuilder(str);
StringBuilder reversed = sb.reverse();
return reversed.toString();
}
public static int reverse(int x) {
int tmp = x;
int oct = 0;
int res = 0;
while (true) {
oct = tmp % 10;
tmp = tmp / 10;
res = (res+oct)*10;
if ((tmp/10) == 0) {
res = res+tmp;
return res;
}
}
}
public static double reverse(int num)
{
double num1 = num;
double ret = 0;
double counter = 0;
while (num1 > 1)
{
counter++;
num1 = num1/10;
}
while(counter >= 0)
{
int lastdigit = num%10;
ret += Math.pow(10, counter-1) * lastdigit;
num = num/10;
counter--;
}
return ret;
}