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。

文章图片
例子：

Input : num = 12345Output : 54321Input : num = 876Output : 678

推荐：请在"实践首先, 在继续解决方案之前。
流程图：

文章图片
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 numdivide by 10 to rev_numrev_num = rev_num*10 + num%10; (b) Divide num by 10(3) Return rev_num

例子：
num = 4562
rev_num = 0
【算法设计（如何编写程序以反转数字（））】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)

#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 = 0while (n > 0 ): a = n % 10 rev = rev * 10 + a n = n / / 10print (rev)# This code is contributed by Shariq Raza

// 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 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# 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：