用Java对数组排序的问题

时间:2019-04-26 13:19:08

标签: java arrays sorting char

我是个流血的初学者,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已删除,但显然我的Arrays.sort()出现错误,但我看不到它。为什么我的Arrays.sort()行中的错误在哪里以及如何解决?

编辑:如果我不这样处理Arrays.sort(),则它可以编译和运行,因此显然这行只有一个问题。如果我将其留在其中,则指向数组并显示错误: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");
    }

}

1 个答案:

答案 0 :(得分:0)

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");
    }       
  }