比较两个词

弗朗切斯科·布鲁(Francesco Brouge)Nutricato

我是该论坛的新手(以注册用户身份),所以我非常努力(我发誓!),不要发布问题,而是寻找旧的答案,并检查了其他人在与我类似的问题中的错误,但是我不能没事。

我的代码在这里,应该检查一个词是否是另一个词的字谜。我很确定自己已经使我的生活变得很复杂,并且会有更简单的方法来完成这项工作,但是...我已经从事此工作了一段时间了,并且希望看到它的工作...

知道为什么不这样做吗?

我所看到的只是空字典,并且当两个单词具有相同的字母数时,这两个词始终是字谜(这意味着实际上我的字典没有做任何事情:'()

import acm.program.ConsoleProgram;
import java.util.*;


public class Anagrams extends ConsoleProgram {

    String firstWord;
    String secondWord;

    public boolean checkLength(String firstWord, String secondWord) {
        if (firstWord.length() == secondWord.length()) {
            println("Same length!");
            return true;
        } else {
            return false;
        }
    }

    public boolean anagram(String firstWord, String secondWord) {
        firstWord = firstWord.toLowerCase();
        secondWord = secondWord.toLowerCase();
        String[] firstArray = firstWord.split("\\a");
        String[] secondArray = secondWord.split("\\a");
        int firstLength = firstWord.length();
        int secondLength = secondWord.length();

        Map<String, Integer> firstDictionary = new HashMap<String, Integer>();
        Map<String, Integer> secondDictionary = new HashMap<String, Integer>();

        for (firstLength = 0; firstLength == firstArray.length; firstLength++) {
            System.out.println("checking the letter " + firstArray[firstLength] + " in array" + firstArray.toString());
            if (firstDictionary.get(firstArray[firstLength]) == null) {
                firstDictionary.put(firstArray[firstLength], 1);
            } else {
                firstDictionary.put(firstArray[firstLength], firstDictionary.get(firstArray[firstLength]) + 1);
            }
        }

        for (secondLength = 0; secondLength == secondArray.length; secondLength++) {
            if (secondDictionary.get(secondArray[secondLength]) == null) {
                secondDictionary.put(secondArray[secondLength], 1);
            } else {
                secondDictionary.put(secondArray[secondLength], secondDictionary.get(secondArray[secondLength]) + 1);
            }
        }

        if (firstDictionary.equals(secondDictionary)) {
            return true;
        } else {
            return false;
        }
    }

    public void run() {
        int runAgain = 0;
        while (runAgain == 0) {
            println("Enter the first word to be analyzed");
            firstWord = readLine();
            println("Enter the second word to be analyzed");
            secondWord = readLine();

            if (checkLength(firstWord, secondWord) == true) {

                if (anagram(firstWord, secondWord) == true) {
                    println("Yes! The two words are anagrams!");
                }
            } else {
                println("No. The two words are not anagrams!");
            }
        }

    }
}
范·邓

您的循环条件有误

  for (firstLength=0;firstLength==firstArray.length;firstLength++)

这应该是

   for (firstLength=0;firstLength<firstArray.length;firstLength++)

由于条件firstLength==firstArray.length为假,因此您的程序将永远不会进入循环。

我相信:

String[] firstArray = firstWord.split("\\a");
String[] secondArray = secondWord.split("\\a");

是错误的,您只需要char数组,因此可以使用以下数组:

 char[] first = firstWord.toCharArray();
 char[] second = secondWord.toCharArray();

工作版本:

 public boolean anagram(String firstWord, String secondWord) {
        firstWord = firstWord.toLowerCase();
        secondWord = secondWord.toLowerCase();
        char[] firstArray = firstWord.toCharArray();
        char[] secondArray = secondWord.toCharArray();
        int firstLength = firstWord.length();
        int secondLength = secondWord.length();

        Map<Character, Integer> firstDictionary = new HashMap<>();
        Map<Character, Integer> secondDictionary = new HashMap<>();

        for (firstLength = 0; firstLength < firstArray.length; firstLength++) {
           // System.out.println("checking the letter " + firstArray[firstLength] + " in array" + firstArray.toString());
            if (!firstDictionary.containsKey(firstArray[firstLength])) {
                firstDictionary.put(firstArray[firstLength], 1);
            } else {
                firstDictionary.put(firstArray[firstLength], firstDictionary.get(firstArray[firstLength]) + 1);
            }
        }

        for (secondLength = 0; secondLength < secondArray.length; secondLength++) {
            if (!secondDictionary.containsKey(secondArray[secondLength])) {
                secondDictionary.put(secondArray[secondLength], 1);
            } else {
                secondDictionary.put(secondArray[secondLength], secondDictionary.get(secondArray[secondLength]) + 1);
            }
        }

        if (firstDictionary.equals(secondDictionary)) {
            return true;
        } else {
            return false;
        }
    }

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