[Java] I implemented the combination.

I tried to implement a combination of math in Java

I implemented in Java a combination that takes different r out of n different. I've seen implementations that loop as many as r, I thought about various things so that it would be okay if r pieces changed.

Source code

The code below is designed to take 3 out of 5.

import java.util.ArrayList;
import java.util.List;

public class Trial {

    public static void main(String[] args) {
        Trial me = new Trial();
        me.exec(args);
    }

    public void exec(String[] args) {
        //Array of strings
        String[] strArray = new String[] { "a", "b", "c", "d", "e" };
        //Get the combination
        String[][] combinations = getCombinations(strArray, 3);
        //Result output
        for (String[] combination : combinations) {
            for (String str : combination) {
                System.out.print("  " + str);
            }
            System.out.println();
        }
    }

    private String[][] getCombinations(String[] strArray, int selectCount) {
        //List for packing results
        List<String[]> list = new ArrayList<>();
        //I use it a little, so I keep it in a variable
        int len = strArray.length;
        //Find the maximum value in binary from the length of the array.
        int dec = 0;
        for (int i = 0; i < len; i++) {
            dec += Math.pow(2, i);
        }
        //Decrement from the maximum value.
        for (int num = dec; 0 < num; num--) {
            //Make it a character string in binary notation. (0 is not filled)
            String bin = Integer.toBinaryString(num);
            if (!isCombination(bin, selectCount)) {
                //Ignore if the number of 1s does not match the selection.
                continue;
            }
            int j = bin.length() - 1;
            int tmplen = len - bin.length();
            String[] combination = new String[selectCount];
            int idx = selectCount - 1;
            for (int i = len - 1; tmplen <= i; i--) {
                if (bin.charAt(j--) == '1') {
                    combination[idx--] = strArray[i];
                }
            }
            list.add(combination);
        }
        return list.toArray(new String[0][0]);
    }

    private boolean isCombination(String str, int selectCount) {
        int sum = 0;
        for (int i = 0; i < str.length(); i++) {
            sum += Character.getNumericValue(str.charAt(i));
        }
        if (sum == selectCount) {
            return true;
        }
        return false;
    }
}

Execution result

  a  b  c
  a  b  d
  a  b  e
  a  c  d
  a  c  e
  a  d  e
  b  c  d
  b  c  e
  b  d  e
  c  d  e

The combination of 5 to 3 is 5! / (2! * (5-2)!) = 10 Maybe it fits.

Finally

I think there is a lot of room for improvement when considering performance and dealing with large numbers.

that's all