本文概述
假设输入是32位整数, 则编写程序以反转整数。如果反向整数溢出, 则输出-1作为输出。
让我们看看整数位数反转的简单方法.
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;
}
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.
python
# 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
的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
?>
输出如下:
6985
但是, 如果数量很大, 以至于反向溢出, 则输出是一些垃圾值。如果我们使用输入为任意大的数字来运行上面的代码, 请说1000000045, 则输出是一些垃圾值, 例如1105032705或任何其他垃圾值。看到这个用于输出。
如何处理溢出?
想法是存储和的先前值可以存储在变量中, 该变量可以每次检查以查看反向是否溢出。
下面是处理这种情况的实现。
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
输出如下:
Reverse of no. is 54321
WARNING OVERFLOWED!!!
Reverse of no. is 0
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