如何检查是否两个词是字谜

问题描述

我有一个程序，向您展示了两个词是否是彼此的字谜。有迹象表明，将无法正常工作的几个例子，我会AP preciate任何帮助，但如果它不先进，将是巨大的，因为我是第一年的程序员。 校长和theclassroom是彼此的字谜游戏，但是当我变theclassroom到theclafsroom还在说他们是字谜，我究竟做错了什么？

 进口的java.util.ArrayList;
公共类AnagramCheck
{
  公共静态无效的主要（字符串的args []）
  {
      字符串phrase1 =tbeclassroom;
      phrase1 =（phrase1.toLowerCase（））修剪（）。
      的char [] phrase1Arr = phrase1.toCharArray（）;

      字符串phrase2 =校长;
      phrase2 =（phrase2.toLowerCase（））修剪（）。
      ArrayList的＆LT;性格＆GT; phrase2ArrList = convertStringToArraylist（phrase2）;

      如果（phrase1.length（）！= phrase2.length（））
      {
          System.out.print（没有字谜present。）;
      }
      其他
      {
          布尔isFound = TRUE;
          的for（int i = 0; I＆LT; phrase1Arr.length;我++）
          {
              对于（INT J = 0; J＆LT; phrase2ArrList.size（）; J ++）
              {
                  如果（phrase1Arr [我] == phrase2ArrList.get（J））
                  {
                      System.out.print（有一个共同的元素\ñ。）;
                      isFound =;
                      phrase2ArrList.remove（J）;
                  }
              }
              如果（isFound ==假）
              {
                  System.out.print（有没有字谜present。）;
                  返回;
              }
          }
          System.out.printf（％s是％S的字谜，phrase1，phrase2）;
      }
  }

  公共静态的ArrayList＆LT;性格＆GT; convertStringToArraylist（字符串str）{
      ArrayList的＆LT;性格＆GT; charlist中的=新的ArrayList＆LT;性格＆GT;（）;
      的for（int i = 0; I＆LT; str.length（）;我++）{
          charList.add（str.charAt（ⅰ））;
      }
      返回charlist中;
  }
}
 

解决方案

最快的算法是将每26个英文字符映射到唯一的素数。然后计算串的乘积。由算术基本定理，2串是字谜当且仅当他们的产品是相同的。

I have a program that shows you whether two words are anagrams of one another. There are a few examples that will not work properly and I would appreciate any help, although if it were not advanced that would be great, as I am a 1st year programmer. "schoolmaster" and "theclassroom" are anagrams of one another, however when I change "theclassroom" to "theclafsroom" it still says they are anagrams, what am I doing wrong?

import java.util.ArrayList;
public class AnagramCheck
{
  public static void main(String args[])
  {
      String phrase1 = "tbeclassroom";
      phrase1 = (phrase1.toLowerCase()).trim();
      char[] phrase1Arr = phrase1.toCharArray();

      String phrase2 = "schoolmaster";
      phrase2 = (phrase2.toLowerCase()).trim();
      ArrayList<Character> phrase2ArrList = convertStringToArraylist(phrase2);

      if (phrase1.length() != phrase2.length())
      {
          System.out.print("There is no anagram present.");
      }
      else
      {
          boolean isFound = true;
          for (int i=0; i<phrase1Arr.length; i++)
          {
              for(int j = 0; j < phrase2ArrList.size(); j++)
              {
                  if(phrase1Arr[i] == phrase2ArrList.get(j))
                  {
                      System.out.print("There is a common element.\n");
                      isFound = ;
                      phrase2ArrList.remove(j);
                  }
              }
              if(isFound == false)
              {
                  System.out.print("There are no anagrams present.");
                  return;
              }
          }
          System.out.printf("%s is an anagram of %s", phrase1, phrase2);
      }
  }

  public static ArrayList<Character> convertStringToArraylist(String str) {
      ArrayList<Character> charList = new ArrayList<Character>();
      for(int i = 0; i<str.length();i++){
          charList.add(str.charAt(i));
      }
      return charList;
  }
}

解决方案

Fastest algorithm would be to map each of the 26 English characters to a unique prime number. Then calculate the product of the string. By the fundamental theorem of arithmetic, 2 strings are anagrams if and only if their products are the same.

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