java - 如何在没有 toCharArray 方法的情况下比较字谜？

Question

/*
 * 
 * 
 */

import java.util.Scanner;
import java.util.Arrays;


public class Anagram
{
  public static void main(String[] args)
  {

Scanner sc = new Scanner(System.in);
String one, two;

System.out.print("Enter first sentence: ");
String s1 = sc.nextLine();
System.out.print("Enter second sentence: ");
String s2 = sc.nextLine();

sc.close();

s1 = s1.toLowerCase();
char[] chars = s1.toCharArray();
Arrays.sort(chars);
String sort1 = new String(chars);
System.out.println(sort1 + " are the letters of " + s1 + " in order");

s2 = s2.toLowerCase();
char[] chars2 = s2.toCharArray();
Arrays.sort(chars2);
String sort2 = new String(chars2);
System.out.println(sort2 + " are the letters of " + s2 + " in order");

if(sort1.equals(sort2))
  System.out.println(s1 + " is an anagram of " + s2);


  }
}

这是我的程序，使用 toCharArray 比较变位词可以正常工作，但是是否可以完成每个 'a' 到 'z' 并将其附加到排序列表而不是 toCharArray？

Answer 1

好吧，有几个选项。如果您只想避免调用“toCharArray”，那么您可以循环遍历字符串并以这种方式创建字符，但我怀疑这就是您要找的东西？

你也可以做如下实现(伪代码):

public void areAnagrams(String s1, String s2)
{
  int[] aNumLetters = new int[26];

  s1.toLowerCase();
  s2.toLowerCase();

  for each char c in s1
    aNumLetters[(int)c - ((int)'a')]++;

  for each char c in s2
    aNumLetters[(int)c - ((int)'a')]--;

  for each int nLetterCount in aNumLetters
    if nLetterCount != 0
      return false

  return true;
}