LeetCode 179: Largest Number

Given a list of non negative integers, arrange them such that they form the largest number.
For example, given [3, 30, 34, 5, 9], the largest formed number is 9534330.
Note: The result may be very large, so you need to return a string instead of an integer.

public class Solution {
    public String largestNumber(int[] num) {
        // Adopt modified LSD (Least-Significant-Digits) algorithm
        // MSD algorithm is not suitable here
        // However, LSD requires identical length of every element (String format)
        // Suppose that the longest length of element is 4, we append any element by itself until the length is 4*2:
        // "123" ---> "123"+"123"+"12" ---> "12312312"
        // We can define a "charAt()" function to implement that without actually append the string.
        int n = num.length;
        int R = 10; // Only 10 digits: 0, 1, 2, ..., 9
        int len = 0; // Max length of elements
        String[] a = new String[n];
        String[] aux = new String[n];
        
        // Convert 'num' to 'a'
        for (int i = 0; i < n; i++)
        {
            a[i] = String.valueOf(num[i]);
            
            if (len < a[i].length())
                len = a[i].length();
        }
        
        // Original num: [3, 30, 34, 5, 9]
        // Result: [30, 3, 34, 5, 9]
        for (int d = len*2-1; d >= 0; d--)
        {
            int[] count = new int[R+1];
            
            // Compute frequency counts
            for (int i = 0; i < n; i++)
                count[charAt(a[i], d)+1]++;
                
            // Transform counts to indices
            for (int r = 0; r < R; r++)
                count[r+1] += count[r];
                
            // Distribute the data
            for (int i = 0; i < n; i++)
                aux[count[charAt(a[i], d)]++] = a[i];
                
            // Copy back
            for (int i = 0; i < n; i++)
                a[i] = aux[i];
        }
        
        String result = "";
        
        for (int i = n-1; i >= 0; i--)
            result += a[i];
        
        // Note: Maybe exist special case: "0000"    
        if (result.length()>1 && result.charAt(0)=='0')
            result = "0";

        return result;
    }
    
    private int charAt(String s, int d)
    {
        d = d%s.length();
        
        return s.charAt(d)-'0';
    }
}