编写程序以反转数字

编写程序以反转整数。

例子 ：

Input : num = 12345
Output: 54321

Input : num = 876
Output: 678

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 = 2650 + 4 = 2654
num = num / 10 = 0

程序：

C++

#include 
 
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 
 
/* 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

PHP

 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
?>

Javascript

    let num = 4562;
    // Function to reverse digits of num
    function reversDigits(num) {
        let rev_num = 0;
        while(num > 0)
        {
            rev_num = rev_num * 10 + num % 10;
            num = Math.floor(num / 10);
        }
        return rev_num;
    }
 // function call  
    document.write(reversDigits(num));
     
// This code is contributed by Surbhi tyagi
 

输出

Reverse of no. is 2654

时间复杂度： O(log(n))，其中n是输入数字。

递归方式

C++

// C++ program to reverse digits of a number
#include 
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 ;
 
/* 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

 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
?>

Java脚本

 
// Javascript program to reverse digits of a number
 
/* Recursive function to reverse digits of num*/
var rev_num = 0;
var base_pos = 1;
function reversDigits(num)
{
 
    if(num > 0)
    {
        reversDigits(Math.floor(num/10));
        rev_num += (num%10)*base_pos;
        base_pos *= 10;
    }
    return rev_num;
}
 
// Driver Code
    let num = 4562;
    document.write("Reverse of no. is "
        + reversDigits(num));
 
// This code is contributed
// by Mayank Tyagi
 
 

输出

Reverse of no. is 2654

时间复杂度： O(log(n))，其中n是输入数字。

在Java使用String

我们将数字转换为在此之后使用StringBuffer的字符串，我们将使用反向扭转的字符串()方法

角壳

Input: 32100

So for the above input if we try to solve this by reversing the string, then the output will be 00123.

So to deal with this situation we again need to convert the string to integer so that our output will be 123

Java

// Java program to reverse a number
 
public class GFG {
    static int reversDigits(int num)
    {
        // converting number to string
        StringBuffer string
            = new StringBuffer(String.valueOf(num));
 
        // reversing the string
        string.reverse();
 
        // converting string to integer
        num = Integer.parseInt(String.valueOf(string));
 
        // returning integer
        return num;
    }
    public static void main(String[] args)
    {
        int num = 4562;
        System.out.println("Reverse of no. is "
                           + reversDigits(num));
    }
}

输出

Reverse of no. is 2654

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个程序，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。