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enter image description hereI am trying to solve this question:
a) Write a method with the following header that takes an integer n and
returns the value of n! (pronounced n factorial) computed as follows:
public static int factorial(int n)
Note that 0! = 1 and n! = n * (n-1) * (n-2)*.....*1.
Example: factorial(4) will return 24 which is = 4*3*2*1.
b) Write a method with the following header that takes an integer x and
returns true if x is a Strong number. Otherwise, it returns false.
public static boolean isStrongNumber(int x)
Note that the isStrongNumber method should call the factorial method to compute the factorial of
each digit in x.
public static int factorial(int n) {
int f =1;
for (int i = 1; i <=n; i++)
f=f*i;
return f;
}
public static boolean isStrongNumber(int x) {
int temp = x;
int z;
int q = 0;
int sum = 0;
while (temp > 0) {
x = x % 10;
z = factorial(x);
q += z;
if (q == temp) {
System.out.print(q + " ");
return true;
}
}
}
This is my answer, but I get an error every time I try to run it.
You did not return boolean value at end of the isStrongNumber method
public static int factorial(int n) {
int result = 1;
for (int i = 2; i <= n; i++) {
result *= i;
}
return result;
}
public static boolean isStrongNumber(int num) {
int originalNum = num;
int sum = 0;
while (num > 0) {
sum += factorial(num % 10);
num /= 10;
}
return sum == originalNum;
}
, main function
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.print("Enter a positive integer: ");
int number = Integer.parseInt(scanner.nextLine());
Set<Integer> set = new TreeSet<>();
for (int i = 1; i <= number; i++) {
if (isStrongNumber(i)) {
set.add(i);
}
}
System.out.println("The Strong numbers between 1 and " + number + " are:");
System.out.println(set);
scanner.close();
}
, output for input 100000
Enter a positive integer: 100000
The Strong numbers between 1 and 100000 are:
[1, 2, 145, 40585]
This cannot compile as it lacks a return statement outside the while loop. In fact, you cant be sure to go inside the loop even once if x<=0 for exmaple. You should add return false outside the loop at the end of the method. Also if you get an error and write a question on StackOverflow, copy the error message it's very helpful.
I tried to use some codes on Stack Overflow but when I used some codes,I didn't get the answer I want, this is my code now:
public class JavaApplication7 {
public static void main(String[] args) {
boolean answer = rev(1221);
if(answer == true) System.out.println("true");
else System.out.println("false");
}
static boolean rev(int number){
int reverse = 0, remain;
int num = number;
while(number > 0){
remain = number % 10;
reverse = (reverse * 10) + remain;
num = num / 10;
}
return number == reverse;
}
}
I want to check if palindrome is equal to the orginal int.
You have two bugs relating to confusing num (the local copy which you're modifying) and number (the parameter which you don't modify). I would make it final to make clear that the value is constant. And then use num consistently. I would also simplify the code a little. Like,
static boolean rev(final int number) {
int reverse = 0;
int num = number;
while (num > 0) {
reverse *= 10;
reverse += num % 10;
num /= 10;
}
return number == reverse;
}
And similarly in main, using an if to print true or false is pointless (and if you're going to do that use if (answer) - don't add unnecessary == true). But, I would prefer,
public static void main(String[] args) {
boolean answer = rev(1221);
System.out.println(answer);
}
Which outputs
true
Your rev function should be as below.
reversedInteger = 0, remainder, originalInteger;
originalInteger = num;
while( num != 0 )
{
remainder = num % 10;
reversedInteger = reversedInteger * 10 + remainder;
num /= 10;
}
static boolean rev(int number){
int reverse = 0;
int num = number;
while(number > 0){
reverse = reverse * 10 + number % 10;
number /= 10;
}
return reverse == num;
}
This is code to find armstorm numbers between given intervals. But when I put my my method in loop in main method so that it can run between given interval, then the loop only run one time and don't change the value passed to method. Why this is so? Is there is a difference in working of these conditions in different languages?
import java.util.Scanner;
class ArmstormNumbers {
int mod, div, count = 0, rev = 0, pow = 1, sum = 0;
int checkArm(int num) {
while(num!=0) {
mod = num%10;
div = num/10;
num = div;
count++;
rev = (rev*10) + mod;
}
while(rev!=0) {
mod = rev%10;
div = rev/10;
rev = div;
int temp = mod;
for(pow = 1; pow < count; pow++) {
mod = mod * temp;
}
sum=sum+(mod);
}
return sum;
}
}
class HelloWorld {
public static void main(String args[]) {
int num, arms, inp, fp, asm, num2;
Scanner input = new Scanner(System.in);
ArmstormNumbers object = new ArmstormNumbers();
System.out.println("This program will find armstorm numbers between two intervels");
System.out.println("Input Initial point ");
num = input.nextInt();
System.out.println("Input Final Point");
num2 = input.nextInt();
int temp = num;
for(num = temp; num <= num2; num++) {
asm = object.checkArm(num);
if(asm == num) {
System.out.println(num);
}
}
}
}
Move the variable declarations so that they become method-local variables which get initialised each time the method is executed, instead of only once when instantiating the object.
I.e. change
int mod,div,count=0,rev=0,pow=1,sum=0;
int CheckArm(int num)
{
to
int CheckArm(int num)
{
int mod,div,count=0,rev=0,pow=1,sum=0;
The problem you have and which you solve that way is probably relying on sum being 0 at start but not making that sure.
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;
}
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;
}