我该如何解决这个猪拉丁语翻译器？

Question

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

public class PigLatin 
{ 



    public static void main (String[] args) { 
    System.out.println("Translate or Reverse"); 
    Scanner choose = new Scanner(System.in); 
    String choice = choose.nextLine(); 
    Scanner input = new Scanner(System.in); 
    String ipWord = input.nextLine(); 




    if (choice.equals("Translate")) 
    { 


     System.out.println(translator(ipWord)); 

    } 
    else if(choice.equals("Reverse")) 
    { 
     System.out.println(reverse(ipWord)); 
    } 
    else 
    { 
     System.out.println("Error"); 
    } 

} 
    public static String translator(String ipWord) 

    { 
    String[] wordsArray = ipWord.split(" "); 
    String result = ""; 

    for(int j =0;j<wordsArray.length;j++) 
    { 

     if(wordsArray[j].charAt(0)=='a' || wordsArray[j].charAt(0) =='e' || wordsArray[j].charAt(0)== 'i' || wordsArray[j].charAt(0)== 'o' || wordsArray[j].charAt(0)== 'u'|| wordsArray[j].charAt(0)=='A' || wordsArray[j].charAt(0) =='E' || wordsArray[j].charAt(0)== 'I' || wordsArray[j].charAt(0)== 'O' || wordsArray[j].charAt(0)== 'U') 
/* I check if it is begin with vowel. If it begin with vowels + "yay". For example "Eat">"Eatyay"*/ 
     { 
     result = result+ " " + wordsArray[j] +"yay"; 
     } 

     else 
     { 
     for (int i = 1; i<= wordsArray.length; i++) 
     { 
      if(wordsArray[j].charAt(0)=='B'||wordsArray[j].charAt(0)=='C'||wordsArray[j].charAt(0)=='D'||wordsArray[j].charAt(0)=='F'||wordsArray[j].charAt(0)=='G'||wordsArray[j].charAt(0)=='H'||wordsArray[j].charAt(0)=='J'||wordsArray[j].charAt(0)=='K'||wordsArray[j].charAt(0)=='L'||wordsArray[j].charAt(0)=='M'||wordsArray[j].charAt(0)=='N'||wordsArray[j].charAt(0)=='P'||wordsArray[j].charAt(0)=='Q'||wordsArray[j].charAt(0)=='R'||wordsArray[j].charAt(0)=='S'||wordsArray[j].charAt(0)=='T'||wordsArray[j].charAt(0)=='V'||wordsArray[j].charAt(0)=='X'||wordsArray[j].charAt(0)=='Z'||wordsArray[j].charAt(0)=='W'||wordsArray[j].charAt(0)=='Y') 
/* I check if it begin with a UpperCase Consonants, then I will take every consonant before the first vowel, move it to the back and add "ay" 
. For example, "World" > "Orldway" ; "String" > "Ingstryay" */ 
      { 
      if (wordsArray[j].charAt(i)=='a' || wordsArray[j].charAt(i) =='e' || wordsArray[j].charAt(i)== 'i' || wordsArray[j].charAt(i)== 'o' || wordsArray[j].charAt(i)== 'u') 
      { 
       result = result+ " " + wordsArray[j].substring(i,i+1).toUpperCase()+ wordsArray[j].substring(i+1)+wordsArray[j].substring(i-1,i).toLowerCase() +"ay"; 
      } 


      } 

      else 
      { 
      if (wordsArray[j].charAt(i)=='a' || wordsArray[j].charAt(i) =='e' || wordsArray[j].charAt(i)== 'i' || wordsArray[j].charAt(i)== 'o' || wordsArray[j].charAt(i)== 'u'|| wordsArray[j].charAt(j)=='A' || wordsArray[j].charAt(j) =='E' || wordsArray[j].charAt(j)== 'I' || wordsArray[j].charAt(j)== 'O' || wordsArray[j].charAt(j)== 'U') 
      { 
       result = result+ " " + wordsArray[j].substring(i)+ wordsArray[j].substring(0,i) +"ay"; 
      } 

      } 



     } 


     } 

    } 
    return result; 
    } 
    public static String reverse (String ipWord) 
    { 
    String[] wordsArray = ipWord.split(" "); 
    String result = ""; 
    for(int h = 0 ; h < wordsArray.length ; h++) 
    { 
     if(wordsArray[h].charAt(wordsArray[h].length()-3)=='y') 

     { 
     result = result + " " + wordsArray[h].charAt(wordsArray[h].length()-4) + wordsArray[h].substring(0,wordsArray[h].length()-4); 

     } 
     else 
     { 
     result = result + wordsArray[h].substring(0,wordsArray[h].length()-2); 
     } 
    } 
    return result; 
    } 
} 

在这里，我修好了。我认为它不能翻译超过2个单词。如果我尝试翻译嗨，你好吗。错误是：

java.lang.StringIndexOutOfBoundsException: String index out of range: 2 
    at java.lang.String.charAt(Unknown Source) 
    at PigLatin.translator(PigLatin.java:56) 
    at PigLatin.main(PigLatin.java:23) 
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method) 
    at sun.reflect.NativeMethodAccessorImpl.invoke(Unknown Source) 
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(Unknown Source) 
    at java.lang.reflect.Method.invoke(Unknown Source) 
    at edu.rice.cs.drjava.model.compiler.JavacCompiler.runCommand(JavacCompiler.java:272)*/ 

Answer 1

始终信任错误。他们知道他们在说什么。如果他们说你超出范围 - 你确实超出范围，你甚至知道确切的位置。您没有标出该行，但不难发现：

if (wordsArray[j].charAt(i)=='a' || ... 

什么可能超出范围？ wordsArray[j]不存在或charAt(i)不存在。这是什么？

j受单词数量的限制。我们在这里很清楚。这留下了另一个。现在，这是什么意思i ...

for (int i = 1; i<= wordsArray.length; i++) 

这似乎也由wordsArray单词量的约束。从1开始，所以它永远不会显示第一个单词，但这是不相关的，因为获取单词不是你使用它的目的。实际上，导致错误的原因是，您希望第j个单词具有与输入中的单词相同的字母。