生成数字的所有旋转

给定整数n ，任务是生成所有可能的左移数。左移数字是当该数字的所有数字都向左移动一个位置，而第一个位置的数字移动到最后一个数字时生成的数字。

例子：

Input: n = 123
Output: 231 312

Input: n = 1445
Output: 4451 4514 5144

方法：

• 假设n ＝ 123 。
• 将n乘以10，即n = n * 10 = 1230 。
• 将第一个数字加到结果数字上，即1230 +1 = 1231 。
• 从结果数中减去(第一个数字)* 10 ^k ，其中k是原始数字的位数(在这种情况下，k = 3)。
• 1231 – 1000 = 231是原始编号的左移编号。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the count of digits of n
int numberOfDigits(int n)
{
    int cnt = 0;
    while (n > 0) {
        cnt++;
        n /= 10;
    }
    return cnt;
}
  
// Function to print the left shift numbers
void cal(int num)
{
    int digits = numberOfDigits(num);
    int powTen = pow(10, digits - 1);
  
    for (int i = 0; i < digits - 1; i++) {
  
        int firstDigit = num / powTen;
  
        // Formula to calculate left shift
        // from previous number
        int left
            = ((num * 10) + firstDigit)
              - (firstDigit * powTen * 10);
        cout << left << " ";
  
        // Update the original number
        num = left;
    }
}
  
// Driver Code
int main()
{
    int num = 1445;
    cal(num);
    return 0;
}

Java

// Java implementation of the approach
class GFG 
{
  
// Function to return the count of digits of n
static int numberOfDigits(int n) 
{
    int cnt = 0;
    while (n > 0) 
    {
        cnt++;
        n /= 10;
    }
    return cnt;
}
  
// Function to print the left shift numbers
static void cal(int num)
{
    int digits = numberOfDigits(num);
    int powTen = (int) Math.pow(10, digits - 1);
  
    for (int i = 0; i < digits - 1; i++)
    {
        int firstDigit = num / powTen;
  
        // Formula to calculate left shift
        // from previous number
        int left = ((num * 10) + firstDigit) -
                    (firstDigit * powTen * 10);
                  
        System.out.print(left + " ");
                  
        // Update the original number
        num = left;
    }
}
  
// Driver Code
public static void main(String[] args) 
{
    int num = 1445;
    cal(num);
}
}
  
// This code is contributed by 
// PrinciRaj1992

Python3

# Python3 implementation of the approach
  
# function to return the count of digit of n
def numberofDigits(n):
    cnt = 0
    while n > 0:
        cnt += 1
        n //= 10
    return cnt
      
# function to print the left shift numbers
def cal(num):
    digit = numberofDigits(num)
    powTen = pow(10, digit - 1)
      
    for i in range(digit - 1):
          
        firstDigit = num // powTen
          
        # formula to calculate left shift 
        # from previous number
        left = (num * 10 + firstDigit - 
               (firstDigit * powTen * 10))
        print(left, end = " ")
          
        # Update the original number
        num = left
          
# Driver code
num = 1445
cal(num)
  
# This code is contributed
# by Mohit Kumar

C#

// C# implementation of the approach
using System;
  
public class GFG{
      
// Function to return the count of digits of n
static int numberOfDigits(int n)
{
    int cnt = 0;
    while (n > 0) {
        cnt++;
        n /= 10;
    }
    return cnt;
}
  
// Function to print the left shift numbers
static void cal(int num)
{
    int digits = numberOfDigits(num);
    int powTen = (int)Math.Pow(10, digits - 1);
  
    for (int i = 0; i < digits - 1; i++) {
  
        int firstDigit = num / powTen;
  
        // Formula to calculate left shift
        // from previous number
        int left
            = ((num * 10) + firstDigit)
            - (firstDigit * powTen * 10);
        Console.Write(left +  " ");
  
        // Update the original number
        num = left;
    }
}
  
// Driver Code
    static public void Main (){
        int num = 1445;
        cal(num);
    } 
}
  
// This code is contributed by akt_mit....

PHP

 0) 
    { 
        $cnt++; 
        $n = floor($n / 10); 
    } 
    return $cnt; 
} 
  
// Function to print the left shift numbers 
function cal($num) 
{ 
    $digits = numberOfDigits($num); 
    $powTen = pow(10, $digits - 1); 
  
    for ($i = 0; $i < $digits - 1; $i++)
    { 
  
        $firstDigit = floor($num / $powTen); 
  
        // Formula to calculate left shift 
        // from previous number 
        $left
            = (($num * 10) + $firstDigit) - 
               ($firstDigit * $powTen * 10); 
              
        echo $left, " "; 
  
        // Update the original number 
        $num = $left; 
    } 
} 
  
// Driver Code
$num = 1445;
cal($num); 
  
// This code is contributed by Ryuga
?>

输出：

4451 4514 5144