Cara menggunakan TYPECASTED pada Python

Menentukan Jenis Variabel

Table of Contents Show

python program for reverse of a number using type casting
How to reverse a number (optimum solution) in Python
How do you reverse a float number?
How do I reverse a number in Python?
What is __ reversed __ in Python?
How do you reverse a 4 digit number in Python?

Mungkin ada saat-saat ketika kita ingin menentukan tipe pada variabel. Sebenarnya hal ini bisa dilakukan menggunakan casting. Python adalah bahasa berorientasi objek, dan karena itu ia menggunakan kelas untuk mendefinisikan tipe data, termasuk tipe primitifnya.

Karena itu, casting dalam python dilakukan menggunakan fungsi konstruktor:

int () – membuat bilangan bulat dari literal integer, literal float (dengan membulatkan ke bilangan bulat sebelumnya), atau string literal (memberikan string mewakili bilangan bulat)
float () – membuat angka float dari literal integer, literal float atau literal string (asalkan string mewakili float atau integer)
str () – membuat string dari berbagai tipe data, termasuk string, literal integer, dan literal float

Contoh:
Integer

x = int(1)
y = int(2.8)
z = int("3")
print(x)
print(y)
print(z)

Float

x = float(1)
y = float(2.8)
z = float("3")
w = float("4.2")
print(x)
print(y)
print(z)
print(w)

String

x = str("s1")
y = str(2)
z = str(3.0)
print(x)
print(y)
print(z)

python program for reverse of a number using type casting

In this tutorial, we are going to learn how to reverse a given number within two lines of code in Python using the type casting method.

Table of Contents

python program for reverse of a number using type casting
How to reverse a number (optimum solution) in Python
How do you reverse a float number?
How do I reverse a number in Python?
What is __ reversed __ in Python?
How do you reverse a 4 digit number in Python?

How to reverse a number (optimum solution) in Python

Before jump into the program 1st, you have some knowledge of typecasting in Python.

following are the way of typecasting in python:

let’s consider two variable

x=’123′ (string variable)
y=123 (integer variable)

1)integer to string:
—-> z=str(y)
—-> o/p= ‘123’

2)string to an integer:
—-> z=int(x)
—-> o/p=123

3)integer to float:
—-> z=float(y)
—-> o/p=123.0

In similar way in this program we are going to use type casting .

Now move on the program.
first, take a value from the user:

#take integer as a string input from users.
x=input("Please enter a integer ")

Now use type cast concept to convert a string because the reverse of number is easier than the reverse of the string so we directly taking a number as a string then reverse them.

the reverse of given number which is taken by the user in a string format.

#reverse of string instead of integer.
y=x[::-1]

Now again typecast the string into integer and print it as an output.

#type cast from string to integer.
y=int(y)
print("Reverse of given number = ",y)

let’s combine the whole code together :

# Reverse of a number in python within very less line of code.
#take integer as a string input from users.
x=input("Please enter a integer ")
#reverse of string instead of integer.
y=x[::-1]
#type cast from string to integer.
y=int(y)
print("Reverse of given number = ",y)

Output:

Please enter a integer 156
Reverse of given number =  651

You may also read:

Python Program to reverse digits of a given number

Write a program to reverse the digits of an integer.

Examples :

Input : num = 12345
Output: 54321

Input : num = 876
Output: 678

Flowchart:

ITERATIVE WAY

Algorithm:

Input:  num
(1) Initialize rev_num = 0
(2) Loop while num > 0
     (a) Multiply rev_num by 10 and add remainder of num  
          divide by 10 to rev_num
               rev_num = rev_num*10 + num%10;
     (b) Divide num by 10
(3) Return rev_num

Example:

num = 4562
rev_num = 0
rev_num = rev_num *10 + num%10 = 2
num = num/10 = 456
rev_num = rev_num *10 + num%10 = 20 + 6 = 26
num = num/10 = 45
rev_num = rev_num *10 + num%10 = 260 + 5 = 265
num = num/10 = 4
rev_num = rev_num *10 + num%10 = 2650 + 4 = 2654
num = num/10 = 0

Program:

C++

#include <bits/stdc++.h>

using namespace std;

int reverseDigits(int num)

{

int rev_num = 0;

while (num > 0) {

rev_num = rev_num * 10 + num % 10;

num = num / 10;

}

return rev_num;

}

int main()

{

int num = 4562;

cout << "Reverse of no. is " << reverseDigits(num);

getchar();

return 0;

}

C

#include <stdio.h>

int reverseDigits(int num)

{

int rev_num = 0;

while (num > 0) {

rev_num = rev_num * 10 + num % 10;

num = num / 10;

}

return rev_num;

}

int main()

{

int num = 4562;

printf("Reverse of no. is %d", reverseDigits(num));

getchar();

return 0;

}

Java

class GFG {

static int reverseDigits(int num)

{

int rev_num = 0;

while (num > 0) {

rev_num = rev_num * 10 + num % 10;

num = num / 10;

}

return rev_num;

}

public static void main(String[] args)

{

int num = 4562;

System.out.println("Reverse of no. is "

+ reverseDigits(num));

}

Python

n = 4562

rev = 0

while(n > 0):

a = n % 10

rev = rev * 10 + a

n = n // 10

print(rev)

C#

using System;

class GFG {

static int reverseDigits(int num)

{

int rev_num = 0;

while (num > 0) {

rev_num = rev_num * 10 + num % 10;

num = num / 10;

}

return rev_num;

}

public static void Main()

{

int num = 4562;

Console.Write("Reverse of no. is "

+ reverseDigits(num));

}

PHP

<?php

function reverseDigits($num)

{

$rev_num = 0;

while($num > 1)

{

$rev_num = $rev_num * 10 +

$num % 10;

$num = (int)$num / 10;

}

return $rev_num;

}

$num = 4562;

echo "Reverse of no. is ",

reverseDigits($num);

?>

Javascript

<script>

let num = 4562;

function reverseDigits(num) {

let rev_num = 0;

while(num > 0)

{

rev_num = rev_num * 10 + num % 10;

num = Math.floor(num / 10);

}

return rev_num;

}

document.write(reverseDigits(num));

</script>

Output

Reverse of no. is 2654

Time Complexity: O(n), where n is the input number.

Auxiliary Space: O(1)

RECURSIVE WAY :

C++

#include <bits/stdc++.h>

using namespace std;

int reverseDigits(int num)

{

static int rev_num = 0;

static int base_pos = 1;

if (num > 0) {

reverseDigits(num / 10);

rev_num += (num % 10) * base_pos;

base_pos *= 10;

}

return rev_num;

}

int main()

{

int num = 4562;

cout << "Reverse of no. is " << reverseDigits(num);

return 0;

}

C

#include <stdio.h>;

int reversDigits(int num)

{

static int rev_num = 0;

static int base_pos = 1;

if (num > 0) {

reversDigits(num / 10);

rev_num += (num % 10) * base_pos;

base_pos *= 10;

}

return rev_num;

}

int main()

{

int num = 4562;

printf("Reverse of no. is %d", reversDigits(num));

getchar();

return 0;

}

Java

class GFG {

static int rev_num = 0;

static int base_pos = 1;

static int reversDigits(int num)

{

if (num > 0) {

reversDigits(num / 10);

rev_num += (num % 10) * base_pos;

base_pos *= 10;

}

return rev_num;

}

public static void main(String[] args)

{

int num = 4562;

System.out.println(reversDigits(num));

}

Python3

rev_num = 0

base_pos = 1

def reversDigits(num):

global rev_num

global base_pos

if(num > 0):

reversDigits((int)(num / 10))

rev_num += (num % 10) * base_pos

base_pos *= 10

return rev_num

num = 4562

print("Reverse of no. is ",

reversDigits(num))

C#

using System;

class GFG {

static int rev_num = 0;

static int base_pos = 1;

static int reversDigits(int num)

{

if (num > 0) {

reversDigits(num / 10);

rev_num += (num % 10) * base_pos;

base_pos *= 10;

}

return rev_num;

}

public static void Main()

{

int num = 4562;

Console.WriteLine(reversDigits(num));

}

PHP

<?php

$rev_num = 0;

$base_pos = 1;

function reversDigits($num)

{

global $rev_num;

global $base_pos;

if($num > 0)

{

reversDigits((int)($num / 10));

$rev_num += ($num % 10) *

$base_pos;

$base_pos *= 10;

}

return $rev_num;

}

$num = 4562;

echo "Reverse of no. is ",

reversDigits($num);

?>

Javascript

<script>

var rev_num = 0;

var base_pos = 1;

function reversDigits(num)

{

if(num > 0)

{

reversDigits(Math.floor(num/10));

rev_num += (num%10)*base_pos;

base_pos *= 10;

}

return rev_num;

}

let num = 4562;

document.write("Reverse of no. is "

+ reversDigits(num));

</script>

Time Complexity: O(log(n))

Auxiliary Space: O(log(n)), where n is the input number.

Using Recursion, without extra variable

C++

#include <iostream>

#include <math.h>

using namespace std;

int len(int number)

{

return log10(number)+1;

}

int rev_number(int& number)

{

if ( (number % 10) == number )

return number;

int last = number % 10;

int remaining = number / 10;

int l = len(remaining);

return last*pow(10, l) + rev_number(remaining);

}

int main()

{

int number = 123456;

cout << rev_number(number) << endl;

return 0;

}

Java

import java.lang.Math;

import java.util.*;

class GFG {

public static int len(int number)

{

return (int)(Math.log10(number)) + 1;

}

public static int rev_number(int number)

{

if (number % 10 == number) {

return number;

}

int last = number % 10;

int remaining = number / 10;

int l = len(remaining);

return last * (int)Math.pow(10, l)

+ rev_number(remaining);

}

public static void main(String[] args)

{

int number = 123456;

System.out.println(rev_number(number));

}

Python3

def number_length(num):

return len(str(num))

def rev_number(number):

if (number % 10) == number:

return number

last = number % 10

remaining = number // 10

l = number_length(remaining)

return last*pow(10, l) + rev_number(remaining)

def main():

number = 123456

print(rev_number(number))

if __name__ == "__main__":

main()

Using String in java

We will convert the number to a string using StringBuffer after this, we will reverse that string using the reverse() method

corner case

Input: 32100
So for the above input if we try to solve this by reversing the string, then the output will be 00123.
So to deal with this situation we again need to convert the string to integer so that our output will be 123

C++

#include <bits/stdc++.h>

using namespace std;

int reverseDigits(int num)

{

string strin = to_string(num);

reverse(strin.begin(), strin.end());

num = stoi(strin);

return num;

}

int main()

{

int num = 4562;

cout << "Reverse of no. is " << reverseDigits(num);

return 0;

}

Java

public class GFG {

static int reversDigits(int num)

{

StringBuffer string

= new StringBuffer(String.valueOf(num));

string.reverse();

num = Integer.parseInt(String.valueOf(string));

return num;

}

public static void main(String[] args)

{

int num = 4562;

System.out.println("Reverse of no. is "

+ reversDigits(num));

}

Python3

def reversDigits(num):

string = str(num)

string = list(string)

string.reverse()

string = ''.join(string)

num = int(string)

return num

if __name__ == "__main__":

num = 4562

print("Reverse of no. is ", reversDigits(num))

C#

using System;

public class GFG{

public static string ReverseString(string s)

{

char[] array = s.ToCharArray();

Array.Reverse(array);

return new string(array);

}

static int reversDigits(int num)

{

string strin = num.ToString();

strin = ReverseString(strin);

num = int.Parse(strin);

return num;

}

static public void Main ()

{

int num = 4562;

Console.Write("Reverse of no. is "

+ reversDigits(num));

}

Javascript

<script>

function reversDigits(num)

{

let str

= num.toString().split("").reverse().join("");

num = parseInt(str);

return str;

}

let num = 4562;

document.write("Reverse of no. is "

+ reversDigits(num));

</script>

Time Complexity: O(log10n) where n is the input number

Auxiliary Space: O(1)

Reverse digits of an integer with overflow handled

Note that the above program doesn’t consider leading zeroes. For example, for 100 programs will print 1. If you want to print 001 then see this comment from Maheshwar.

Using Slicing in Python:

C++

#include <bits/stdc++.h>

using namespace std;

int reversDigits(int num){

string str = to_string(num);

reverse(str.begin(), str.end());

num = stoll(str);

return num;

}

int main()

{

int num = 4562;

cout << "Reverse of no. is " << reversDigits(num) ;

return 0;

}

Python3

def reversDigits(num):

string = str(num)

string = string[::-1]

num = int(string)

return num

if __name__ == "__main__":

num = 4562

print("Reverse of no. is ", reversDigits(num))

C#

using System;

class GFG {

static int reversDigits(int num)

{

string str = Convert.ToString(num);

char[] charArray = str.ToCharArray();

Array.Reverse(charArray);

str = new string(charArray);

num = Convert.ToInt32(str);

return num;

}

public static void Main(string[] args)

{

int num = 4562;

Console.Write("Reverse of no. is "

+ reversDigits(num));

}

Javascript

function reversDigits(num){

let str = num.toString().split("");

str.reverse();

num = parseInt(str.join(""))

return num;

}

let num = 4562;

console.log("Reverse of no. is " + reversDigits(num));

Java

import java.util.*;

class GFG {

static int reversDigits(int num)

{

String str = String.valueOf(num);

str = new StringBuilder(str).reverse().toString();

num = Integer.valueOf(str);

return num;

}

public static void main(String[] args)

{

int num = 4562;

System.out.println("Reverse of no. is "

+ reversDigits(num));

}

Time Complexity: O(n), where n is the input number

Auxiliary Space: O(1)

Try extensions of above functions that should also work for floating-point numbers.

How do you reverse a float number?

Convert the float into an integer and devisor model, reverse the integer part using standard number reversing also and then apply devisor..

Eg: 3.14159..

Get decimal location by chceking the number this number is lessthan 3 so exponent can be 10^1..

If a float is 956.123 then exponent would be 10^3..

Reverse the int -> 951413..

How do I reverse a number in Python?

Algorithm.

Input Integer: number..

(1) Initialize variable revs_number = 0..

(2) Loop while number > 0..

(a) Multiply revs_number by 10 and add remainder of number..

divide by 10 to revs_number..

revs_number = revs_number*10 + number%10;.

(b) Divide num by 10..

(3) Return revs_number..

What is reversed in Python?

Python reversed() method The reversed() method returns the reversed iterator of the given sequence. It is the same as the iter() method but in reverse order. Internally, it calls the __reversed__() method of the sequence class.

How do you reverse a 4 digit number in Python?

In this program, while loop is used to reverse a number as given in the following steps: First, the remainder of the num divided by 10 is stored in the variable digit . Now, the digit contains the last digit of num , i.e. 4. digit is then added to the variable reversed after multiplying it by 10.