Java中数组排序的问题

Question

我是一个该死的初学者，试图编写一个 lil 程序来检查 2 个单词是否是字谜。到目前为止，单词中的所有空格都被删除了，但显然我的 Arrays.sort() 有错误，但我看不到它。为什么以及在我的 Arrays.sort() 行中出现错误，我该如何解决？

编辑：如果我像这样离开 Arrays.sort() ，它会编译并工作，所以显然该行只有一个问题。如果我把它们留在它指向数组并说错误：找不到符号

public static void isAnagramm(String wordOne, String wordTwo)       
{

    String  w1= wordOne.replaceAll("\\s", ""); 
    int word1 = w1.length();
    String w2 = wordTwo.replaceAll("\\s", "");
    int word2 = w2.length();

    boolean anagrammStatus = false;



    if(word1 == word2)
    {
        anagrammStatus = true;
    }
    else
    {
        char [] charArrayWordOne = w1.toLowerCase().toCharArray(); 
        char [] charArrayWordTwo = w2.toLowerCase().toCharArray();  

        //Arrays.sort(charArrayWordOne); 
        //Arrays.sort(charArrayWordTwo);

        anagrammStatus = charArrayWordOne.equals(charArrayWordTwo);

    }

    if(anagrammStatus == false)
    {
        System.out.println("Anagram");
    }                   
    else;
    {
        System.out.println("No Anagram");
    }

}

Answer 1

这应该可以解决问题：

  public static void isAnagramm(String wordOne, String wordTwo)       
  {   
    String w1= wordOne.replaceAll("\\s", "");
    String w2 = wordTwo.replaceAll("\\s", "");

    // No need to keep the length variables

    boolean anagramStatus = false;

    // Check if the strings are equal to begin with, use equals and not  == operator
    if(w1.equals(w2)) 
    {
      anagramStatus = true;
    }
    else
    {
      char [] charArrayWordOne = w1.toLowerCase().toCharArray();
      char [] charArrayWordTwo = w2.toLowerCase().toCharArray();  

      Arrays.sort(charArrayWordOne);
      Arrays.sort(charArrayWordTwo);

      // Compare arrays using the Arrays.equals method to avoid comparing the object references
      anagramStatus = Arrays.equals(charArrayWordOne, charArrayWordTwo);
    }

    // Use simple boolean logic in your condition here, or again, always use == instead of =
    if (anagramStatus) 
    {
      System.out.println("Anagram");
    }                   
    else
    {
      System.out.println("No Anagram");
    }       
  }