算法设计：如何编写程序以反转数字？

本文概述

• 建议：在继续解决方案之前, 请先在"实践"上解决它。
• C ++
• C
• Java
• python
• C#
• 的PHP
• C ++
• C
• Java
• Python3
• C#
• 的PHP

编写一个程序以反转整数。

例子 ：

Input : num = 12345
Output : 54321

Input : num = 876
Output : 678

推荐：请在"实践首先, 在继续解决方案之前。

流程图：

迭代方式

算法：

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

例子：

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 = 265 + 4 = 2654

num = num / 10 = 0

程序：

C ++

#include <bits/stdc++.h>
  
using namespace std;
/* Iterative function to reverse digits of num*/
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 = 4562;
     cout << "Reverse of no. is "
          << reversDigits(num);
     getchar ();
     return 0;
}
  
// This code is contributed
// by Akanksha Rai(Abby_akku)

C

#include <stdio.h>
  
/* Iterative function to reverse digits of num*/
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 = 4562;
     printf ( "Reverse of no. is %d" , reversDigits(num));
  
     getchar ();
     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
?>

时间复杂度：

O(Log(n))其中n是输入数字。

输出如下：

2654

递归方式

感谢Raj将其添加到原始帖子中。

C ++

// C++ program to reverse digits of a number
#include <bits/stdc++.h>
using namespace std;
/* Recursive function to reverse digits of num*/
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;
}
  
// Driver Code
int main()
{
     int num = 4562;
     cout << "Reverse of no. is " 
          << reversDigits(num);
  
     return 0;
}
  
// This code is contributed 
// by Akanksha Rai(Abby_akku)

C

// C program to reverse digits of a number
#include <stdio.h>;
  
/* Recursive function to reverse digits of num*/
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;
}
  
/*Driver program to test reversDigits*/
int main()
{
     int num = 4562;
     printf ( "Reverse of no. is %d" , reversDigits(num));
  
     getchar ();
     return 0;
}

Java

// Java program to reverse digits of a number
  
// Recursive function to 
// reverse digits of num
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;
}
  
// Driver Code
public static void main(String[] args)
{
     int num = 4562 ;
     System.out.println(reversDigits(num));
}
}
  
// This code is contributed by mits

Python3

# Python 3 program to reverse digits
# of a number
rev_num = 0
base_pos = 1
  
# Recursive function to reverse
# digits of num
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
      
# Driver Code 
num = 4562
print ( "Reverse of no. is " , reversDigits(num))
      
# This code is contributed by Rajput-Ji

C#

// C# program to reverse digits of a number
  
// Recursive function to 
// reverse digits of num
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;
}
  
// Driver Code
public static void Main()
{
     int num = 4562;
     Console.WriteLine(reversDigits(num));
}
}
  
// This code is contributed 
// by inder_verma

的PHP

<?php
// PHP program to reverse digits of a number
$rev_num = 0; 
$base_pos = 1;
  
/* Recursive function to 
reverse digits of num*/
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 ; 
} 
  
// Driver Code
$num = 4562; 
echo "Reverse of no. is " , reversDigits( $num ); 
  
// This code is contributed by ajit
?>

输出如下：

Reverse of no. is 2654

时间复杂度：O(Log(n))其中n是输入数字。

整数的倒数位, 已处理溢出

请注意, 上述程序不考虑前导零。例如, 对于100程序将打印1。如果要打印001, 请参阅Maheshwar的注释。

尝试上述功能的扩展, 这些扩展也应适用于浮点数。