通过连接数组旋转中的每个元素的最大数量

给定一个包含 N 个元素的数组。任务是通过连接每个旋转中的每个元素来打印最大数量。在每次旋转中，第一个元素将取代每个旋转中的最后一个元素，反之亦然。
例子：

Input: a[]: {54, 546, 548, 60} 
Output: 6054546548 
1st Rotation: 5465486054 
2nd Rotation: 5486054546 
3rd Rotation: 6054546548 
4th Rotation: 5454654860
Input: a[]: {1, 4, 18, 96} 
Output: 961418

处理方法：仔细观察，发现所有元素中最左边一位数最大的数将是该数中的第一个元素。由于串联必须根据数组的旋转来完成。连接从最大的最左边数字索引到末尾的所有数字，然后连接从^第0^个索引到最大最左边数字索引的元素。
下面是上述方法的实现：

C++

// C++ program to print the
// Maximum number by concatenating
// every element in rotation of array
#include 
using namespace std;
 
// Function to print the largest number
void printLargest(int a[], int n)
{
 
    // store the index of largest
    // left most digit of elements
    int max = -1;
    int ind = -1;
 
    // Iterate for all numbers
    for (int i = 0; i < n; i++) {
 
        int num = a[i];
 
        // check for the last digit
        while (num) {
            int r = num % 10;
            num = num / 10;
            if (num == 0) {
                // check for the largest left most digit
                if (max < r) {
                    max = r;
                    ind = i;
                }
            }
        }
    }
 
    // print the largest number
 
    // print the rotation of array
    for (int i = ind; i < n; i++)
        cout << a[i];
 
    // print the rotation of array
    for (int i = 0; i < ind; i++)
        cout << a[i];
}
 
// Driver Code
int main()
{
    int a[] = { 54, 546, 548, 60 };
    int n = sizeof(a) / sizeof(a[0]);
    printLargest(a, n);
    return 0;
}

Java

// Java program to print the
// Maximum number by concatenating
// every element in rotation of array
import java.util.*;
import java.lang.*;
 
public class GFG {
    // Function to print the largest number
    static void printLargest(int a[], int n)
    {
        // store the index of largest
        // left most digit of elements
        int max = -1;
        int ind = -1;
 
        // Iterate for all numbers
        for (int i = 0; i < n; i++) {
            int num = a[i];
 
            // check for the last digit
            while (num > 0) {
                int r = num % 10;
                num = num / 10;
                if (num == 0) {
                    // check for the largest left most digit
                    if (max < r) {
                        max = r;
                        ind = i;
                    }
                }
            }
        }
        // print the largest number
 
        // print the rotation of array
        for (int i = ind; i < n; i++)
            System.out.print(a[i]);
 
        // print the rotation of array
        for (int i = 0; i < ind; i++)
            System.out.print(a[i]);
    }
 
    // Driver Code
    public static void main(String args[])
    {
        int a[] = { 54, 546, 548, 60 };
        int n = a.length;
        printLargest(a, n);
    }
}

Python3

# Python program to print the
# Maximum number by concatenating
# every element in rotation of array
  
# Function to print the largest number
def printLargest(a, n):
 
      
    # store the index of largest
    # left most digit of elements
    max =-1
    ind =-1
      
    # Iterate for all numbers
    for i in range(0, n):
         num = a[i]
          
        # check for the last digit
         while(num):
         
            r = num % 10;
            num = num / 10;
            if(num == 0):
                # check for the largest left most digit
                if(max

---

C#

// C# program to print the
// Maximum number by concatenating
// every element in rotation of array
using System;
 
class GFG {
 
    // Function to print the largest number
    static void printLargest(int[] a, int n)
    {
        // store the index of largest
        // left most digit of elements
        int max = -1;
        int ind = -1;
 
        // Iterate for all numbers
        for (int i = 0; i < n; i++) {
            int num = a[i];
 
            // check for the last digit
            while (num > 0) {
                int r = num % 10;
                num = num / 10;
                if (num == 0) {
                    // check for the largest left most digit
                    if (max < r) {
                        max = r;
                        ind = i;
                    }
                }
            }
        }
        // print the largest number
 
        // print the rotation of array
        for (int i = ind; i < n; i++)
            Console.Write(a[i]);
 
        // print the rotation of array
        for (int i = 0; i < ind; i++)
            Console.Write(a[i]);
    }
 
    // Driver Code
    public static void Main()
    {
        int[] a = { 54, 546, 548, 60 };
        int n = 4;
        printLargest(a, n);
    }
}
 
// This code is contributed by mohit kumar 29

Javascript

 
// Javascript program to print the
// Maximum number by concatenating
// every element in rotation of array
 
    // Function to prlet the largest number
    function printLargest(a, n)
    {
        // store the index of largest
        // left most digit of elements
        let max = -1;
        let ind = -1;
   
        // Iterate for all numbers
        for (let i = 0; i < n; i++) {
            let num = a[i];
   
            // check for the last digit
            while (num > 0) {
                let r = num % 10;
                num = Math.floor(num / 10);
                if (num == 0) {
                    // check for the largest
                    // left most digit
                    if (max < r) {
                        max = r;
                        ind = i;
                    }
                }
            }
        }
        // print the largest number
   
        // print the rotation of array
        for (let i = ind; i < n; i++)
            document.write(a[i]);
   
        // print the rotation of array
        for (let i = 0; i < ind; i++)
            document.write(a[i]);
    }
 
// driver code
 
     let a = [ 54, 546, 548, 60 ];
        let n = a.length;
        printLargest(a, n);
   

输出：

6054546548

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