Java递归电话号码信件 [英] Java recursion phone number letters

问题描述

如何使用递归方法编写java程序，该方法接收类似234的int并将其转换为电话簿上的相应字母（2 = ABC，3 = DEF等），并打印出来这种排列？例如：

How do you write a java program using a recursive method that takes in an int like "234" and converts this into the corresponding letters on a phone pad (2 = ABC, 3 = DEF, etc), and prints out the permutations of this? e.g.:

输入= 234

输出= ADG ADH ADI AEG AEH AEI AFG AFH AFI BDG BDH BDI BEG BEH BEI BFG BFH BFI CDG CDH CDI CEG CEH CEI CFG CFH CFI

output = ADG ADH ADI AEG AEH AEI AFG AFH AFI BDG BDH BDI BEG BEH BEI BFG BFH BFI CDG CDH CDI CEG CEH CEI CFG CFH CFI

input = 89

input = 89

输出= TW TX TY TZ UW UX UY UZ VW VX VY VZ

output = TW TX TY TZ UW UX UY UZ VW VX VY VZ

推荐答案

这是一个版本没有数组或arraylist。结果将根据您的要求打印到标准输出。

Here's a version without either array or arraylist. Results are printed to stdout as you requested.

String[] allLetters = new String[] {
                "0",
                "1",
                "ABC",
                "DEF",
                "GHI",
                "JKL",
                // etc...
        };

public static void convert(String phoneNumber)
{
  convertSubstring(phoneNumber,"");
}

private static void convertSubstring(String phoneNumber, String convertedLetters)
{                   
  int digit = Integer.parseInt(phoneNumber.substring(0, 1));
  String letters=allLetters[digit];
  String remainingString=phoneNumber.substring(1);

  for (int i = 0; i < letters.length(); ++i) 
  {
     char letter = letters.charAt(i);
     String result=convertedLetters+letter;
     if (remainingString.length()==0)
        System.out.println(result);
     else
        convertSubstring(remainingString, result);
  }
}

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