代從Java中的字符串數組

Question

您好那裏的所有編碼字符串中的字符，任何幫助表示讚賞，這是一個個人項目，而不是爲我以爲會很有趣的一所學校，但沒有談到了將所以。現在該項目是讀入一個輸入文件，將其與一個srting []字母= {「a」，「b」，「c」....「z」}進行比較，然後用另一個字符串替換字母[ ] newAlphabets = {「蘋果」，「球」，「貓」，....「斑馬」}。現在我已經嘗試了簡單的replaceAll（）示例，它有點工作。但計劃是輸入一個.txt文件，並用新文件進行替換運行並輸出。 JamesBond.txt：姓名是Bond，James Bond。 Output.txt的應該是：NoAppleMonkeyElephant InsectSnake BallOpenNoDuck，JungelAppleMonkeyElephantSnake BallOpenNoDuck。 這是我到目前爲止有：

import java.io.IOException; 
import java.io.File; 
import java.util.Scanner; 

public class FnR { 
// array of alphabets 
static String[] alphabet = {"a","b","c","d","e","f","g","h","i", 
     "j","k","l","m","n","o","p","q","r","s","t","u", 
     "v","w","x","y","z"}; 
// replace the alphabets with substituted string 
static String[] newAlphabets = 
{"Apple","Bat","Cat","Dog","Egale","Fox","Goat","Horse","Insect","Jungle", 
"King","Lion","Monkey","Nose","Open","Push","Quit","Run","Stop","Turn", 
"Up","Volume","Water","X-mas","Yes","Zip"}; 
@SuppressWarnings("resource") 
public static void main(String[] arg) throws IOException{ 
    try { 
     //reads in the file 
     Scanner input = new Scanner("JamesBond.txt"); 
     File file = new File(input.nextLine()); 
     input = new Scanner(file); 

     while (input.hasNextLine()) { 
      String line = input.nextLine(); 
      //prints the txt file out 
      System.out.println(line); 
      // replaces the letter j with Jungel in the text regardless of 
     case 
      String newtxt1 = line.replaceAll("J","Jungle"); 

      //prints new text 
      System.out.println(newtxt1); 
     } 
     input.close(); 
    } catch (Exception ex) { 
     ex.printStackTrace(); 
    } 
} } 
Output: 
Name is Bond, James Bond 
Name is Bond, Jungleames Bond 

Answer 1

不會給你的代碼。這個答案提供了一些提示，讓你自己編寫代碼。

請勿使用replaceAll()。你必須做26次，這不是一個很好的解決方案。

相反，建立結果字符串創建StringBuilder。然後使用字符串方法length()和charAt()以正常的for循環遍歷輸入字符串的字符。

現在，這裏是簡化代碼的主要「技巧」。 Java以Unicode存儲文本，其中連續存儲字母a至z。這意味着你可以用計算這個索引到你的newAlphabets數組中。要做到這一點，你會寫這樣的代碼：

char ch = ...; 
if (ch >= 'a' && ch <= 'z') { 
    int idx = ch - 'a'; // number between 0 and 25, inclusive 
    // use idx here 
} else if (ch >= 'A' && ch <= 'Z') { 
    int idx = ch - 'A'; // number between 0 and 25, inclusive 
    // use idx here 
} else { 
    // not a letter 
} 

希望這可以幫助你自己寫代碼。