比较数组中的元素以查找对，但是在找到两个对时无法弄清楚该怎么做

Question

I am trying to compare elements in a single array to pull out any pairs, three of a kind, four of a kind, or five of a kind... I can't seem to wrap my head around how to do so. I also have to print out the number in which is in a pair, three of a kind, etc... This is what I have so far. Any clues or ideas?

import java.util.Arrays;

public class DieGame3 {
    public static void main(String[] args) {
        Die die1 = new Die(6);

        System.out.println("Welcome to the pairing game!");

        System.out.print("The 5 rolled dice: ");
        int[] numbers = new int[5];
        for(int i = 0; i < numbers.length; i++) {
            die1.roll();
            numbers[i] = die1.getValue();
        }
        System.out.println(Arrays.toString(numbers));
        findPairs(numbers);
    }
    public static int[] findPairs(int[] d1) {
        int count = 0;
        for (int i = 0; i < d1.length; i++) {
            for (int k = i + 1; k < d1.length; k++) {
                if (d1[i] == d1[k]) {
                    count++;
                }
            }
        }
            if(count == 1)
                System.out.println("You've got " + count + " pair.");
            else
                System.out.println("You've got " + count + " of a kind.");
        return d1;
        }
    }

Answer 1

The key is to somehow keep track of which elements in the array have already been counted. The simplest method is to use a boolean array of the same length, and set it to true when you find a matching element.

public static void findPairs(int[] d1) {  
  boolean[] seen = new boolean[d1.length];
  for (int i = 0; i < d1.length-1; i++) {
    if(!seen[i])
    {
      int count = 1;
      for (int k = i + 1; k < d1.length; k++) {
          if (d1[i] == d1[k]) {
            seen[k] = true;
            count++;
          }
      }
      if(count > 1)
        System.out.printf("%d %ds\n", count, d1[i]);
    }
  }
}

When called with input [5, 5, 3, 3, 3] this prints

2 5s
3 3s