The most efficient way to reverse a number

Question

I am looking for an efficient algorithm to reverse a number, e.g.

Input: 3456789

Output: 9876543

In C++ there are p

Answer 1

Something like this will work:

#include <iostream>

int main()
{
    long in = 3456789;
    long out = 0;
    while(in)
    {
        out *= 10;
        out += in % 10;
        in /= 10;
    }
    std::cout << out << std::endl;
    return 0;
}

Answer 2

You may convert the number to string and then reverse the string with STL algorithms. Code below should work:

 long number = 123456789;
 stringstream ss;
 ss << number;
 string numberToStr = ss.str();

 std::reverse(numberToStr.begin(), numberToStr.end());

 cout << atol(numberToStr.c_str());

You may need to include those relevant header files. I am not sure whether it is the most efficient way, but STL algorithms are generally very efficient.

Answer 3

 //Recursive method to find the reverse of a number  
 #include <bits/stdc++.h> 
    using namespace std; 
    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; 
        cout << "Reverse  " << reversDigits(num); 
    } ``

Answer 4

#include <stdio.h>
unsigned int reverse(unsigned int val)
{
 unsigned int retval = 0;

 while( val > 0)
 {
     retval  = 10*retval + val%10;
     val     /= 10;
 }
 printf("returning - %d", retval);
 return retval;
}


int main()
{
    reverse(123);
}

Answer 5

static public int getReverseInt(int value) {
    int resultNumber = 0;
    for (int i = value; i != 0;) {
        int d = i / 10;
        resultNumber = (resultNumber - d) * 10 + i;
        i = d;
    }
    return resultNumber;
}

I think this will be the fastest possible method without using asm. Note that d*10 + i is equivalent to i%10 but much faster since modulo is around 10 times slower than multiplication.
I tested it and it is about 25 % faster than other answers.

Answer 6

This is the easiest one:

#include<iostream>
using namespace std;

int main()
{
int number, reversed=0;
cout<<"Input a number to Reverse: ";
cin>>number;
    while(number!=0)
  {
    reversed= reversed*10;
    reversed=reversed+number%10;
    number=number/10;
  }
cout<<"Reversed number is: "<<reversed<<endl;

}