将字符串中的字符按字母顺序排序，没有数组或映射（Java）

Question

I am trying to make a program which tests if two separate phrases are anagrams of each other. To do this, I want to sort each string phrase into alphabetical order and compare the results. However, I want to do this without using any arrays or maps . So far I have some pretty messy code/pseudocode:

import java.util.Scanner;

public class AnagramComparer{
  public static void main(String[] args){
    Scanner scan = new Scanner(System.in);

    System.out.println("Enter a sentence.");

    String sentence = scan.nextLine();    

    sentence = sentence.toLowerCase();

    String empty1 = "";

    for(int i = 0; i <= sentence.length(); i++){

      //if char is between 97 & 122

      // if char i <= charAt (0 to sentence length - x)               

      if(sentence.charAt(i) <= empty1.charAt(0)){
        empty1.replace(empty1.charAt(0), sentence.charAt(i));                   
      }
      else{ 
        empty1 = (empty1 + (sentence.charAt(i)));
      }
    } 
    System.out.println(empty1);
    System.out.println(sentence);
  }
}

The idea here is that I would run through the string sentence character-by-character and sort it alphabetically in the string empty1. Feel free to tell me if I'm going in the wrong direction; any help would be much appreciated.

Answer 1

Strings are immutable, so empty1.replace returns the new string, it doesn't modify the existing one.

Since String uses a char[] internally, a new String means a new char[] , so you're still using arrays . You might as well use your own char[] for better performance and lower memory footprint.

char[] chars = sentence.toLowerCase().toCharArray();
Arrays.sort(chars);

Answer 2

Well, arranging characters in a string into alphabetical order is a good idea, but it could be better to make a string like: String alph = "abcdefghijklmnopqrstuvwxyz" and check that the count of every character in alph in both strings are the same.

Something like:

String alph = "abcdefghijklmnopqrstuvwxyz"
for (int i = 0; i < alph.length(); i++){
    if (StringUtils.countMatches(string1, alph.charAt(i) != StringUtils.countMatches(string2, alph.charAt(i)){
        // not an anagram
    }
}