Programmieren lernen

def umkehrung(zahl):
   s=str(zahl)
   z=int(s) 
   s=''

   while (z>=1):
      s=s+str(z%10)
      z = z /10

   print s

umkehrung(124234)
umkehrung('123443434')
                

public class Main {

    public static void main(String[] args) {
        int num = 1030;
        System.out.println( reverse1( num ) );
        System.out.println( reverse2( num ) );
    }
    
    public static int reverse1( int z ) {
        StringBuffer s = new StringBuffer( ).append( z ).reverse();

        return Integer.parseInt( s.toString() );
    }

    public static int reverse2( int z ) {
        int ziffer, ret = 0;

        while ( z != 0 ) {
            ziffer = z % 10;
            ret *= 10;
            ret += ziffer;
            z /= 10;
        }
        return ret;
    }
}

                

a = raw_input("Zahl = ")[::-1]
while a.startswith("0"):
    a = a[1:]
print a
                

Lösung von: Name nicht veröffentlicht

x = int(input('Zahl: '))
u = 0
while x > 0:
    u = u * 10 + x % 10
    x = x // 10
print(u)

                

Lösung von: Claude Vonlanthen (Kantonsschule Olten)

x = str(int(input('Zahl: ')))
print(x[::-1])
                

Lösung von: Claude Vonlanthen (Kantonsschule Olten)

function reverse(num) {
  return parseInt(num.toString().split('').reverse().join(''));
}

console.log(reverse(2454));
console.log(reverse(20010));                                 // lissalanda@gmx.at
                

Lösung von: Lisa Salander (Heidi-Klum-Gymnasium Bottrop)

// NET 6.x | C# 10.x | VS-2022

// Variante 1
static int ReverseInt1(int n) => int.Parse(string.Join("", n.ToString().Reverse()));

// Variante 2
static int ReverseInt2(int n) => Convert.ToInt32(new string(n.ToString().Reverse().ToArray()));

// Variante 3
static int ReverseInt3(int n) {
    var m = 0;
    while (n > 0) {
        m = n % 10 + m * 10;
        n /= 10;
    }
    return m;
}

// Variante 4
static int ReverseInt4(int n) {
    var s = n.ToString().ToCharArray();
    for (int i = 0; i < s.Length / 2; i++)
        (s[i], s[^(i + 1)]) = (s[^(i + 1)], s[i]);
    return int.Parse(s);
}
                

Lösung von: Jens Kelm (@JKooP)

// C++ 14 | VS-2022
#include <iostream>
#include <string>
#include <algorithm>

const auto reverse_int_1(int n) {
    auto str{ std::to_string(n) };
    std::reverse(str.begin(), str.end());
    return std::stoi(str);
}

const auto reverse_int_2(int n) {
    auto m{ 0 };
    while (n) {
        m = n % 10 + m * 10;
        n /= 10;
    }
    return m;
}

const auto reverse_int_3(int n) {
    auto str{ std::to_string(n) };
    for (size_t i{ 0 }; i < str.length() / 2; ++i)
        std::swap(str[i], str[str.length() - i - 1]);
    return std::stoi(str);
}

const auto reverse_int_4(int n) {
    std::string out{};
    auto str{ std::to_string(n) };
    for (auto it{ str.rbegin() }; it != str.rend(); ++it) {
        out.push_back(*it);
    }
    return std::stoi(out);
}

const auto reverse_int_5(int n) {
    auto str{ std::to_string(n) };
    std::string out{str};
    std::copy(str.rbegin(), str.rend(), out.begin());
    return std::stoi(out);
}
                

Lösung von: Jens Kelm (@JKooP)