我想按字母順序對重復的字符進行排序，我該怎么辦？

Question

import java.util.Scanner;
public class Chupapi {

static Scanner input = new Scanner(System.in);

public static void main(String[] args) {

    String str = "";
    System.out.print("Enter string: ");
    str = input.nextLine();

    getConsonantCount(str);
}

public static void getConsonantCount(String str) {
    char string[] = str.toCharArray();

    System.out.print("Duplicate characters in a given string: ");
    for(int i = 0, count = 0; i <string.length; i++) {
        count = 1;
        for(int j = i+1; j <string.length; j++) {
            if(string[i] == string[j] && string[i] != ' ') {
                count++;
                string[j] = '0';
            }
        }
        if(count > 1 && string[i] != '0')
            System.out.print(string[i]);
    }
  }
}

我想整理出“給定字符串中的重復字符：” output 就像按字母順序的示例一樣，

"Duplicate characters in a given string: heuro"
 I want it to be
"Duplicate characters in a given string: ehoru"

這就是我想用這段代碼做的所有事情，關於我應該做什么或使用什么的任何提示？

Answer 1

在查找重復字符之前，使用Arrays.sort()對數組進行排序。

public static void getConsonantCount(String str)
{
    char string[] = str.toCharArray();

    Arrays.sort(string);
    
    System.out.print("Duplicate characters in a given string: ");
    for (int i = 0, count = 0; i < string.length; i++)
    {
        count = 1;
        for (int j = i + 1; j < string.length; j++)
        {
            if (string[i] == string[j] && string[i] != ' ')
            {
                count++;
                string[j] = '0';
            }
        }
        if (count > 1 && string[i] != '0')
            System.out.print(string[i]);
    }
}