Escriba un programa para invertir un entero suponiendo que la entrada es un entero de 32 bits. Si el entero invertido se desborda, imprima -1 como salida.
Veamos un enfoque simple para invertir los dígitos de un número entero .
C++
// A simple C program to reverse digits of
// an integer.
#include <bits/stdc++.h>
using namespace std;
int reversDigits(int num)
{
int rev_num = 0;
while (num > 0) {
rev_num = rev_num * 10 + num % 10;
num = num / 10;
}
return rev_num;
}
/* Driver program to test reversDigits */
int main()
{
int num = 5896;
cout << "Reverse of no. is " << reversDigits(num);
return 0;
}
// This code is contributed by Sania Kumari Gupta (kriSania804)
C
// A simple C program to reverse digits of
// an integer.
#include <stdio.h>
int reversDigits(int num)
{
int rev_num = 0;
while (num > 0) {
rev_num = rev_num * 10 + num % 10;
num = num / 10;
}
return rev_num;
}
/* Driver program to test reversDigits */
int main()
{
int num = 5896;
printf("Reverse of no. is %d", reversDigits(num));
return 0;
}
// This code is contributed by Sania Kumari Gupta (kriSania804)
Java
// Java program to reverse a number
class GFG
{
/* Iterative function to reverse
digits of num*/
static int reversDigits(int num)
{
int rev_num = 0;
while(num > 0)
{
rev_num = rev_num * 10 + num % 10;
num = num / 10;
}
return rev_num;
}
// Driver code
public static void main (String[] args)
{
int num = 4562;
System.out.println("Reverse of no. is "
+ reversDigits(num));
}
}
// This code is contributed by Anant Agarwal.
Python3
# Python program to reverse a number n = 4562; rev = 0 while(n > 0): a = n % 10 rev = rev * 10 + a n = n // 10 print(rev) # This code is contributed by Shariq Raza
C#
// C# program to reverse a number
using System;
class GFG
{
// Iterative function to
// reverse digits of num
static int reversDigits(int num)
{
int rev_num = 0;
while(num > 0)
{
rev_num = rev_num * 10 + num % 10;
num = num / 10;
}
return rev_num;
}
// Driver code
public static void Main()
{
int num = 4562;
Console.Write("Reverse of no. is "
+ reversDigits(num));
}
}
// This code is contributed by Sam007
PHP
<?php
// Iterative function to
// reverse digits of num
function reversDigits($num)
{
$rev_num = 0;
while($num > 1)
{
$rev_num = $rev_num * 10 +
$num % 10;
$num = (int)$num / 10;
}
return $rev_num;
}
// Driver Code
$num = 4562;
echo "Reverse of no. is ",
reversDigits($num);
// This code is contributed by aj_36
?>
Javascript
<script>
// A simple Javascript program to reverse digits of
// an integer.
function reversDigits(num)
{
let rev_num = 0;
while (num > 0)
{
rev_num = rev_num*10 + num%10;
num = Math.floor(num/10);
}
return rev_num;
}
/* Driver program to test reversDigits */
let num = 5896;
document.write("Reverse of no. is "+ reversDigits(num));
// This code is contributed by Mayank Tyagi
</script>
Producción
Reverse of no. is 6985
Complejidad de tiempo: O(log(num))
Espacio Auxiliar: O(1)
Sin embargo, si el número es tan grande que el reverso se desborda, la salida es un valor basura. Si ejecutamos el código anterior con la entrada como cualquier número grande, digamos 1000000045 , entonces la salida es un valor basura como 1105032705 o cualquier otro valor basura. Ver esto para la salida.
¿Cómo manejar el desbordamiento?
La idea es almacenar el valor anterior de la suma en una variable que se puede verificar cada vez para ver si el reverso se desbordó o no.
A continuación se muestra la implementación para hacer frente a tal situación.
C++
// C++ program to reverse digits
// of a number
#include <bits/stdc++.h>
using namespace std;
/* Iterative function to reverse
digits of num*/
int reversDigits(int num)
{
// Handling negative numbers
bool negativeFlag = false;
if (num < 0)
{
negativeFlag = true;
num = -num ;
}
int prev_rev_num = 0, rev_num = 0;
while (num != 0)
{
int curr_digit = num % 10;
rev_num = (rev_num * 10) + curr_digit;
// checking if the reverse overflowed or not.
// The values of (rev_num - curr_digit)/10 and
// prev_rev_num must be same if there was no
// problem.
if ((rev_num - curr_digit) /
10 != prev_rev_num)
{
cout << "WARNING OVERFLOWED!!!"
<< endl;
return 0;
}
prev_rev_num = rev_num;
num = num / 10;
}
return (negativeFlag == true) ?
-rev_num : rev_num;
}
// Driver Code
int main()
{
int num = 12345;
cout << "Reverse of no. is "
<< reversDigits(num) << endl;
num = 1000000045;
cout << "Reverse of no. is "
<< reversDigits(num) << endl;
return 0;
}
// This code is contributed
// by Akanksha Rai(Abby_akku)
C
// C program to reverse digits of a number
#include <stdio.h>
/* Iterative function to reverse digits of num*/
int reversDigits(int num)
{
// Handling negative numbers
bool negativeFlag = false;
if (num < 0)
{
negativeFlag = true;
num = -num ;
}
int prev_rev_num = 0, rev_num = 0;
while (num != 0)
{
int curr_digit = num%10;
rev_num = (rev_num*10) + curr_digit;
// checking if the reverse overflowed or not.
// The values of (rev_num - curr_digit)/10 and
// prev_rev_num must be same if there was no
// problem.
if ((rev_num - curr_digit)/10 != prev_rev_num)
{
printf("WARNING OVERFLOWED!!!\n");
return 0;
}
prev_rev_num = rev_num;
num = num/10;
}
return (negativeFlag == true)? -rev_num : rev_num;
}
/* Driver program to test reverse Digits */
int main()
{
int num = 12345;
printf("Reverse of no. is %d\n", reversDigits(num));
num = 1000000045;
printf("Reverse of no. is %d\n", reversDigits(num));
return 0;
}
Java
// Java program to reverse digits of a number
class ReverseDigits
{
/* Iterative function to reverse digits of num*/
static int reversDigits(int num)
{
// Handling negative numbers
boolean negativeFlag = false;
if (num < 0)
{
negativeFlag = true;
num = -num ;
}
int prev_rev_num = 0, rev_num = 0;
while (num != 0)
{
int curr_digit = num%10;
rev_num = (rev_num*10) + curr_digit;
// checking if the reverse overflowed or not.
// The values of (rev_num - curr_digit)/10 and
// prev_rev_num must be same if there was no
// problem.
if ((rev_num - curr_digit)/10 != prev_rev_num)
{
System.out.println("WARNING OVERFLOWED!!!");
return 0;
}
prev_rev_num = rev_num;
num = num/10;
}
return (negativeFlag == true)? -rev_num : rev_num;
}
public static void main (String[] args)
{
int num = 12345;
System.out.println("Reverse of no. is " + reversDigits(num));
num = 1000000045;
System.out.println("Reverse of no. is " + reversDigits(num));
}
}
Python3
# Python program to reverse digits
# of a number
""" Iterative function to reverse
digits of num"""
def reversDigits(num):
# Handling negative numbers
negativeFlag = False
if (num < 0):
negativeFlag = True
num = -num
prev_rev_num = 0
rev_num = 0
while (num != 0):
curr_digit = num % 10
rev_num = (rev_num * 10) + curr_digit
# checking if the reverse overflowed or not.
# The values of (rev_num - curr_digit)/10 and
# prev_rev_num must be same if there was no
# problem.
if (rev_num >= 2147483647 or
rev_num <= -2147483648):
rev_num = 0
if ((rev_num - curr_digit) // 10 != prev_rev_num):
print("WARNING OVERFLOWED!!!")
return 0
prev_rev_num = rev_num
num = num //10
return -rev_num if (negativeFlag == True) else rev_num
# Driver code
if __name__ =="__main__":
num = 12345
print("Reverse of no. is ",reversDigits(num))
num = 1000000045
print("Reverse of no. is ",reversDigits(num))
# This code is contributed
# Shubham Singh(SHUBHAMSINGH10)
C#
// C# program to reverse digits
// of a number
using System;
class GFG
{
/* Iterative function to reverse
digits of num*/
static int reversDigits(int num)
{
// Handling negative numbers
bool negativeFlag = false;
if (num < 0)
{
negativeFlag = true;
num = -num ;
}
int prev_rev_num = 0, rev_num = 0;
while (num != 0)
{
int curr_digit = num % 10;
rev_num = (rev_num * 10) +
curr_digit;
// checking if the reverse overflowed
// or not. The values of (rev_num -
// curr_digit)/10 and prev_rev_num must
// be same if there was no problem.
if ((rev_num - curr_digit) / 10 != prev_rev_num)
{
Console.WriteLine("WARNING OVERFLOWED!!!");
return 0;
}
prev_rev_num = rev_num;
num = num / 10;
}
return (negativeFlag == true) ?
-rev_num : rev_num;
}
// Driver Code
static public void Main ()
{
int num = 12345;
Console.WriteLine("Reverse of no. is " +
reversDigits(num));
num = 1000000045;
Console.WriteLine("Reverse of no. is " +
reversDigits(num));
}
}
// This code is contributed by ajit
Javascript
<script>
// JavaScript program to reverse digits
// of a number
/* Iterative function to reverse
digits of num*/
function reversDigits(num)
{
// Handling negative numbers
let negativeFlag = false;
if (num < 0)
{
negativeFlag = true;
num = -num ;
}
let prev_rev_num = 0, rev_num = 0;
while (num != 0)
{
let curr_digit = num % 10;
rev_num = (rev_num * 10) + curr_digit;
// checking if the reverse overflowed or not.
// The values of (rev_num - curr_digit)/10 and
// prev_rev_num must be same if there was no
// problem.
if (rev_num >= 2147483647 ||
rev_num <= -2147483648)
rev_num = 0;
if (Math.floor((rev_num - curr_digit) / 10) != prev_rev_num)
{
document.write("WARNING OVERFLOWED!!!"
+ "<br>");
return 0;
}
prev_rev_num = rev_num;
num = Math.floor(num / 10);
}
return (negativeFlag == true) ?
-rev_num : rev_num;
}
// Driver Code
let num = 12345;
document.write("Reverse of no. is "
+ reversDigits(num) + "<br>");
num = 1000000045;
document.write("Reverse of no. is "
+ reversDigits(num) + "<br>");
// This code is contributed by Surbhi Tyagi.
</script>
Producción
Reverse of no. is 54321 WARNING OVERFLOWED!!! Reverse of no. is 0
Complejidad de tiempo: O(log(num))
Espacio Auxiliar: O(1)
Enfoque eficiente:
El enfoque anterior no funcionará si nos dan un entero x de 32 bits con signo, devolvemos x con sus dígitos invertidos. Si invertir x hace que el valor salga del rango de enteros de 32 bits con signo [-231, 231 – 1], devuelva 0.
Entonces no podemos multiplicar el número*10 y luego verificar si el número se desborda o no.
Debemos verificar la condición de desbordamiento antes de multiplicar por 10 usando la siguiente lógica:
Está verificando el caso límite antes de realizar la operación. (invertido > INT_MAX ) no funcionaría
porque invertido se desbordará y se volverá negativo si supera MAX_VALUE.
Dividir MAX_VALUE por 10 le permite verificar la condición sin desbordarse
INT_MAX es igual a 2147483647. INT_MIN es igual a -2147483648.
Los últimos dígitos son 7 y 8. Esta es la razón por la que también verificamos las condiciones rem > 7 y rem < -8
C++
// C++ program to reverse digits
// of a number
#include <bits/stdc++.h>
using namespace std;
int reversDigits(int num) {
int rev = 0 ;
while(num != 0){
int rem = num % 10 ;
num /= 10 ;
if(rev > INT_MAX/10 || rev == INT_MAX/10 && rem > 7){
return 0 ;
}
if(rev < INT_MIN/10 || rev == INT_MIN/10 && rem < -8){
return 0 ;
}
rev = rev*10 + rem ;
}
return rev ;
}
// Driver Code
int main()
{
int num = 12345;
cout << "Reverse of no. is "
<< reversDigits(num) << endl;
num = 1000000045;
cout << "Reverse of no. is "
<< reversDigits(num) << endl;
return 0;
}
Java
// Java program to reverse digits
// of a number
public class GFG
{
static int reversDigits(int num)
{
int rev = 0 ;
while(num != 0){
int rem = num % 10 ;
num /= 10 ;
if(rev > Integer.MAX_VALUE/10 || rev == Integer.MAX_VALUE/10 && rem > 7){
return 0 ;
}
if(rev < Integer.MIN_VALUE/10 || rev == Integer.MIN_VALUE/10 && rem < -8){
return 0 ;
}
rev = rev*10 + rem ;
}
return rev ;
}
// Driver code
public static void main (String[] args)
{
int num = 12345;
System.out.println("Reverse of no. is " + reversDigits(num) );
num = 1000000045;
System.out.println("Reverse of no. is " + reversDigits(num) );
}
}
// This code is contributed by jana_sayantan.
Python3
# Python program for the above approach
INT_MAX = 2147483647
INT_MIN = -2147483648
def reversDigits(num):
rev = 0
while(num > 0):
rem = num % 10
num = (num//10)
if(rev > INT_MAX/10 or rev == INT_MAX/10 and rem > 7):
return 0
if(rev < INT_MIN/10 or rev == INT_MIN/10 and rem < -8):
return 0
rev = rev*10 + rem
return rev
# Driver Code
num = 12345
print(f"Reverse of no. is {reversDigits(num)}")
num = 1000000045
print(f"Reverse of no. is {reversDigits(num)}")
# This code is contributed by shinjanpatra
C#
// C# program to reverse digits
// of a number
using System;
public class GFG
{
static int reversDigits(int num)
{
int rev = 0 ;
while(num != 0){
int rem = num % 10 ;
num /= 10 ;
if(rev > Int32.MaxValue/10 || rev == Int32.MaxValue/10 && rem > 7){
return 0 ;
}
if(rev < Int32.MinValue/10 || rev == Int32.MinValue/10 && rem < -8){
return 0 ;
}
rev = rev*10 + rem ;
}
return rev ;
}
// Driver code
public static void Main (string[] args)
{
int num = 12345;
Console.WriteLine("Reverse of no. is " + reversDigits(num) );
num = 1000000045;
Console.WriteLine("Reverse of no. is " + reversDigits(num) );
}
}
// This code is contributed by phasing17
Javascript
<script>
// JavaScript program to reverse digits
// of a number
const INT_MAX = 2147483647;
const INT_MIN = -2147483648;
function reversDigits(num) {
let rev = 0;
while(num > 0){
let rem = num % 10 ;
num = Math.floor(num/10) ;
if(rev > INT_MAX/10 || rev == INT_MAX/10 && rem > 7){
return 0 ;
}
if(rev < INT_MIN/10 || rev == INT_MIN/10 && rem < -8){
return 0 ;
}
rev = rev*10 + rem ;
}
return rev ;
}
// Driver Code
let num = 12345;
document.write(`Reverse of no. is ${reversDigits(num)}`,"</br>");
num = 1000000045;
document.write(`Reverse of no. is ${reversDigits(num)}`,"</br>");
// This code is contributed by shinjanpatra
</script>
Producción
Reverse of no. is 54321 Reverse of no. is 0
Complejidad de tiempo: O(log(num))
Espacio Auxiliar: O(1)
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Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA