🔗 Arranging Integers in Descending Order

The function solution receives an integer n as a parameter. Return a new integer with the digits of n arranged from largest to smallest. For example, if n is 118372, it should return 873211.

• n is a natural number between 1 and 8000000000, inclusive.

Arrange the integer's digits in descending order. Rather than simply reversing the digits, you need to sort them in descending order by digit value and then return them as a number again.

If you've already solved Reversing a Natural Number into an Array, this one is easy to solve.

Similarly, store each digit into an ArrayList, then sort it with Collections.sort().

Then reassemble the digits into a full number and return it. Be careful not to return it as an int.

JAVA
import java.util.ArrayList;
import java.util.Collections;

/**
 * Arranging Integers in Descending Order class
 *
 * @author RWB
 * @since 2021.12.13 Mon 19:04:28
 */
class Solution
{
	/**
	 * Method that returns the answer
	 *
	 * @param n: [long] natural number
	 *
	 * @return [long] answer
	 */
	public long solution(long n)
	{
		long answer = 0;
		
		ArrayList<Integer> list = new ArrayList<>();
		
		while (n >= 10)
		{
			list.add((int) (n % 10));
			
			n /= 10;
		}
		
		list.add((int) n);
		
		Collections.sort(list);
		
		for (int i = 0; i < list.size(); i++)
		{
			answer += list.get(i) * Math.pow(10, i);
		}
		
		return answer;
	}
}