Related
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.");
}
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;
}
The purpose of my code is to determine the number of times the number 3 appears between a range of numbers, the lower and upper bounds determined by the user.
So far, I can check if the number 3 is in the ten's place my using the modulus. But I am having trouble figuring out if a 3 resides in the hundreds, thousandths, etc. I know I need to use a nested loop, but I can't quite figure out how to code it.
Here is my code so far.
public class JavaThree {
public static void main (String [] args) {
int count = 0;
int num;
System.out.print("Enter lower end: ");
int lowerEnd = IO.readInt();
System.out.print("Enter upper end: ");
int upperEnd = IO.readInt();
if (lowerEnd > upperEnd) {
IO.reportBadInput();
return;
} else {
for(num = lowerEnd; num <= upperEnd; num++) {
if(num % 10 == 3) {
count = count + 1;
} else {
count = count;
}
}
}
IO.outputIntAnswer(count);
}
}
here is proper for loop for your task:
for(num = lowerEnd; num <= upperEnd; num++)
{
int nNum = num;
while (nNum > 0)
{
if( (nNum % 10) == 3)
count = count + 1;
nNum = nNum / 10;
}
}
Another solution, although not as efficient as the solution proposed #Ilya Bursov is to convert the number to a string and count the appearences of the char '3':
int threeCount = 0;
for (int num = lowerEnd; num < upperEnd; num++) {
String strNumber = String.valueOf(num);
for (int i = 0; i < strNumber.length(); i++) {
if (strNumber.charAt(i) == '3') {
threeCount++;
}
}
}
This code gets an integer n and displays all palindrome numbers less than n.
But seems the for loop doesn't work; because when I enter a number except 0, 1 and negatives, nothing happens.
I tried debugging, but couldn't find the problem!
Sample input: 30
Sample output: 1 2 3 4 5 6 7 8 9 11 22
import java.util.Scanner;
public class PalindromeNumbers {
public static void main(String args[]) {
Scanner input = new Scanner(System.in);
int n = input.nextInt();
if (n == 0 || n == 1 || n < 0)
System.out.println("There's no Palindrome Number!");
else {
for (int num = 1; num < n; num++) {
int reversedNum = 0;
int temp = 0;
while (num > 0) {
// use modulus operator to strip off the last digit
temp = num % 10;
// create the reversed number
reversedNum = reversedNum * 10 + temp;
num = num / 10;
}
if (reversedNum == num)
System.out.println(num);
}
}
}
}
You run into an infinite loop: you use num in your for loop as an index, and reset it to 0 inside the loop. Use different variables, and it should work!
for (int i = 1; i < n; i++) {
int num = i;
...
if (reversedNum == i)
System.out.println(i);
}
You can do it in a more concise way:
public static void main(final String args[]) {
final Scanner input = new Scanner(System.in);
final int max = input.nextInt();
if (max <= 0) {
System.out.println("There's no Palindrome Number!");
} else {
for (int i = 1; i < max; i++) {
if (isPalindrome(i)) {
System.out.println(i);
}
}
}
}
private static boolean isPalindrome(final int num) {
final String temp = "" + num;
return new StringBuilder(temp).reverse().toString().equals(temp);
}
You are changing your num variable inside your for loop. The next time num < n is executed, the value changed (to 0). Try something like this:
for (int num = 1; num < n; num++) {
int reversedNum = 0;
int temp = 0;
int temp2 = num;
while (temp2 > 0) {
// use modulus operator to strip off the last digit
temp = temp2 % 10;
// create the reversed number
reversedNum = reversedNum * 10 + temp;
temp2 = temp2 / 10;
}
if (reversedNum == num)
System.out.println(num);
}
This way, you use a temp variable to calculate your reversedNum, and still keeps the value of num for the next loop iteration.
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;
}