旋转给定数字的位数可能达到的最大值

给定正整数N ，任务是在整数N的数字的所有旋转中找到最大值。

例子：

Input: N = 657
Output: 765
Explanation: All rotations of 657 are {657, 576, 765}. The maximum value among all these rotations is 765.

Input: N = 7092
Output: 9270
Explanation:
All rotations of 7092 are {7092, 2709, 9270, 0927}. The maximum value among all these rotations is 9270.

方法：这个想法是找到数字N的所有旋转，并在所有生成的数字中打印最大值。请按照以下步骤解决问题：

• 计算数字N中存在的位数，即对数₁₀ N的上限。
• 使用N的值初始化变量ans ，以存储生成的结果最大数量。
• 遍历范围 [1，记录₁₀ (N)– 1]，然后执行以下步骤：
  • 下一轮更新N的值。
  • 现在，如果生成的下一个旋转超过ans ，则使用N的旋转值更新ans
• 完成上述步骤后，将ans的值打印为所需的答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the maximum value
// possible by rotations of digits of N
void findLargestRotation(int num)
{
    // Store the required result
    int ans = num;
 
    // Store the number of digits
    int len = floor(log10(num) + 1);
 
    int x = pow(10, len - 1);
 
    // Iterate over the range[1, len-1]
    for (int i = 1; i < len; i++) {
 
        // Store the unit's digit
        int lastDigit = num % 10;
 
        // Store the remaining number
        num = num / 10;
 
        // Find the next rotation
        num += (lastDigit * x);
 
        // If the current rotation is
        // greater than the overall
        // answer, then update answer
        if (num > ans) {
            ans = num;
        }
    }
 
    // Print the result
    cout << ans;
}
 
// Driver Code
int main()
{
    int N = 657;
    findLargestRotation(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG
{
 
// Function to find the maximum value
// possible by rotations of digits of N
static void findLargestRotation(int num)
{
   
    // Store the required result
    int ans = num;
 
    // Store the number of digits
    int len = (int)Math.floor(((int)Math.log10(num)) + 1);
    int x = (int)Math.pow(10, len - 1);
 
    // Iterate over the range[1, len-1]
    for (int i = 1; i < len; i++) {
 
        // Store the unit's digit
        int lastDigit = num % 10;
 
        // Store the remaining number
        num = num / 10;
 
        // Find the next rotation
        num += (lastDigit * x);
 
        // If the current rotation is
        // greater than the overall
        // answer, then update answer
        if (num > ans) {
            ans = num;
        }
    }
 
    // Print the result
    System.out.print(ans);
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 657;
    findLargestRotation(N);
}
}
 
// This code is contributed by sanjoy_62.

Python3

# Python program for the above approach
 
# Function to find the maximum value
# possible by rotations of digits of N
def findLargestRotation(num):
   
    # Store the required result
    ans = num
     
    # Store the number of digits
    length = len(str(num))
    x = 10**(length - 1)
     
    # Iterate over the range[1, len-1]
    for i in range(1, length):
       
        # Store the unit's digit
        lastDigit = num % 10
         
        # Store the remaining number
        num = num // 10
         
        # Find the next rotation
        num += (lastDigit * x)
         
        # If the current rotation is
        # greater than the overall
        # answer, then update answer
        if (num > ans):
            ans = num
             
    # Print the result
    print(ans)
 
# Driver Code
N = 657
findLargestRotation(N)
 
# This code is contributed by rohitsingh07052.

C#

// C# program for the above approach
using System;
class GFG{
 
// Function to find the maximum value
// possible by rotations of digits of N
static void findLargestRotation(int num)
{
   
    // Store the required result
    int ans = num;
 
    // Store the number of digits
    double lg = (double)(Math.Log10(num) + 1);
    int len = (int)(Math.Floor(lg));
    int x = (int)Math.Pow(10, len - 1);
 
    // Iterate over the range[1, len-1]
    for (int i = 1; i < len; i++) {
 
        // Store the unit's digit
        int lastDigit = num % 10;
 
        // Store the remaining number
        num = num / 10;
 
        // Find the next rotation
        num += (lastDigit * x);
 
        // If the current rotation is
        // greater than the overall
        // answer, then update answer
        if (num > ans) {
            ans = num;
        }
    }
 
    // Print the result
    Console.Write(ans);
}
 
// Driver Code
public static void Main(string[] args)
{
    int N = 657;
    findLargestRotation(N);
}
}
 
// This code is contributed by souravghosh0416,

Javascript

 
// Javasript program for the above approach
 
// Function to find the maximum value
// possible by rotations of digits of N
function findLargestRotation(num)
{
    // Store the required result
    let ans = num;
 
    // Store the number of digits
    let len = Math.floor(Math.log10(num) + 1);
 
    let x = Math.pow(10, len - 1);
 
    // Iterate over the range[1, len-1]
    for (let i = 1; i < len; i++) {
 
        // Store the unit's digit
        let lastDigit = num % 10;
 
        // Store the remaining number
        num = parseInt(num / 10);
 
        // Find the next rotation
        num += (lastDigit * x);
 
        // If the current rotation is
        // greater than the overall
        // answer, then update answer
        if (num > ans) {
            ans = num;
        }
    }
 
    // Print the result
    document.write(ans);
}
 
// Driver Code
let N = 657;
findLargestRotation(N);
 
// This code is contributed by souravmahato348.

输出：

765

时间复杂度： O(log ₁₀ N)
辅助空间： O(1)