How To Find Length Of Integer In Java
Program to count digits in an integer (4 Different Methods)
Count the number of digits in a long integer entered by a user.
Elementary Iterative Solution
The integer entered by the user is stored in the variable due north. Then the while loop is iterated until the test expression n != 0 is evaluated to 0 (false).
- Later the first iteration, the value of n will be 345 and the count is incremented to 1.
- After the 2d iteration, the value of n will exist 34 and the count is incremented to two.
- Later the third iteration, the value of n will be 3 and the count is incremented to 3.
- At the commencement of the fourth iteration, the value of due north will exist 0 and the loop is terminated.
Then the examination expression is evaluated for false and the loop terminates.
C++
#include <bits/stdc++.h>
using namespace std;
int countDigit( long long northward)
{
int count = 0;
while (northward != 0)
{
n = n / 10;
++count;
}
return count;
}
int primary( void )
{
long long n = 345289467;
cout << "Number of digits : " << countDigit(n);
render 0;
}
C
#include <stdio.h>
int countDigit( long long n)
{
int count = 0;
while (due north != 0)
{
northward = due north / ten;
++count;
}
render count;
}
int master( void )
{
long long n = 345289467;
printf ( "Number of digits : %d" , countDigit(n));
return 0;
}
Java
class GFG {
static int countDigit( long due north)
{
int count = 0 ;
while (north != 0 ) {
north = due north / 10 ;
++count;
}
return count;
}
public static void master(String[] args)
{
long n = 345289467 ;
Organization.out.print( "Number of digits : "
+ countDigit(north));
}
}
Python3
def countDigit(n):
count = 0
while due north ! = 0 :
n / / = 10
count + = one
return count
n = 345289467
impress ( "Number of digits : % d" % (countDigit(n)))
C#
using Organization;
class GFG {
static int countDigit( long n)
{
int count = 0;
while (n != 0)
{
north = n / 10;
++count;
}
return count;
}
public static void Principal()
{
long n = 345289467;
Console.WriteLine( "Number of"
+ " digits : " + countDigit(n));
}
}
PHP
<?php
office countDigit( $n )
{
$count = 0;
while ( $n != 0)
{
$north = round ( $n / 10);
++ $count ;
}
return $count ;
}
$n = 345289467;
echo "Number of digits : "
. countDigit( $n );
?>
Javascript
<script>
function countDigit(n)
{
allow count = 0;
while (n != 0)
{
due north = Math.floor(n / 10);
++count;
}
return count;
}
north = 345289467;
certificate.write( "Number of digits : " + countDigit(n));
</script>
Output
Number of digits : 9
Recursive Solution:
C++
#include <bits/stdc++.h>
using namespace std;
int countDigit( long long n)
{
if (n/x == 0)
return 1;
return 1 + countDigit(n / x);
}
int principal( void )
{
long long north = 345289467;
cout << "Number of digits :" << countDigit(n);
return 0;
}
C
#include <stdio.h>
int countDigit( long long n)
{
if (n/10 == 0)
return 1;
return one + countDigit(north / 10);
}
int master( void )
{
long long n = 345289467;
printf ( "Number of digits : %d" , countDigit(northward));
return 0;
}
Coffee
import java.util.*;
course GFG {
static int countDigit( long n)
{
if (n/ ten == 0 )
return 1 ;
return 1 + countDigit(n / ten );
}
public static void main(Cord[] args)
{
long n = 345289467 ;
Organization.out.print( "Number of digits : "
+ countDigit(n));
}
}
Python3
def countDigit(north):
if north / 10 = = 0 :
render 1
return ane + countDigit(n / / 10 )
n = 345289467
print ( "Number of digits : % d" % (countDigit(n)))
C#
using System;
class GFG {
static int countDigit( long n)
{
if (n/10 == 0)
return one;
return 1 + countDigit(due north / 10);
}
public static void Main()
{
long due north = 345289467;
Panel.WriteLine( "Number of "
+ "digits : "
+ countDigit(n));
}
}
PHP
<?php
part countDigit( $n )
{
if ( $due north /ten == 0)
return 1;
return 1 + countDigit((int)( $n / 10));
}
$due north = 345289467;
print ( "Number of digits : " .
(countDigit( $n )));
?>
Javascript
<script>
function countDigit(northward)
{
if (n/10 == 0)
render i;
return 1 + countDigit(parseInt(n / 10));
}
var n = 345289467;
document.write( "Number of digits :" + countDigit(north));
</script>
Output
Number of digits :9
Log-based Solution:
We can employ log10(logarithm of base of operations ten) to count the number of digits of positive numbers (logarithm is not defined for negative numbers).
Digit count of N = upper bound of log10(N).
C++
#include <bits/stdc++.h>
using namespace std;
int countDigit( long long n) {
render floor ( log10 (n) + 1);
}
int main( void )
{
long long n = 345289467;
cout << "Number of digits : "
<< countDigit(n);
return 0;
}
C
#include <math.h>
#include <stdio.h>
int countDigit( long long n) {
return floor ( log10 (n) + 1);
}
int main( void )
{
long long n = 345289467;
printf ( "Number of digits : %d" , countDigit(n));
return 0;
}
Java
import java.util.*;
course GFG {
static int countDigit( long north)
{
render ( int )Math.floor(Math.log10(n) + 1 );
}
public static void main(Cord[] args)
{
long north = 345289467 ;
System.out.print( "Number of digits : "
+ countDigit(north));
}
}
Python3
import math
def countDigit(northward):
return math.floor(math.log10(n) + 1 )
north = 345289467
print ( "Number of digits : % d" % (countDigit(north)))
C#
using Arrangement;
grade GFG {
static int countDigit( long n)
{
return ( int )Math.Floor(Math.Log10(n) + ane);
}
public static void Main()
{
long n = 345289467;
Console.WriteLine( "Number of digits : "
+ countDigit(n));
}
}
PHP
<?php
function countDigit( $northward )
{
return floor (log10( $n )+1);
}
$n = 345289467;
echo "Number of digits : " ,
countDigit( $n );
?>
Javascript
<script>
function countDigit(north)
{
return Math.floor(Math.log10(north) + 1);
}
var north = 345289467;
document.write( "Number of digits : " +
countDigit(n));
</script>
Output
Number of digits : 9
Method 4:
We can convert the number into a string and so find the length of the cord to get the number of digits in the original number.
C++
#include <$.25/stdc++.h>
using namespace std;
void count_digits( int north)
{
cord num = to_string(n);
cout << num.size() << endl;
}
int main()
{
int north = 345;
count_digits(n);
return 0;
}
Coffee
import coffee.util.*;
public class GFG {
static void count_digits( int north)
{
Cord num = Integer.toString(n);
Organisation.out.println(+num.length());
}
public static void main(String args[])
{
int northward = 345 ;
count_digits(north);
}
}
Python3
def count_digits(n):
n = str (n)
return len (north)
n = 456533457776
print (count_digits(n))
C#
using System;
using Arrangement.Collections.Generic;
course GFG {
static void count_digits( int due north)
{
cord num = Convert.ToString(due north);
Console.WriteLine(+num.Length);
}
public static void Main( cord [] args)
{
int northward = 345;
count_digits(n);
}
}
Javascript
<script>
office count_digits(n)
{
let num = n.toString();
certificate.write(num.length);
}
permit north = 345;
count_digits(n);
</script>
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