题目描述 “为了方便记忆，他通过一种算法把这个密码变换成” 错了吧，明明是用算法把明文变换成密码🙄

解题思路：通过多个分支判断语句将字符对应转换，大小写转换不用查ASCII码表，通过字符的加减就能实现。 

import java.util.Scanner;

public class Main{
    public static void main(String[] args){
        Scanner in = new Scanner(System.in);
        while(in.hasNextLine()){
            String cleartext = in.nextLine();    //读取明文
            
            StringBuilder builder = new StringBuilder();
            for(int i=0; i < cleartext.length(); i++){
                //小写字母转换为数字
                if(cleartext.charAt(i)>='a' && cleartext.charAt(i)<='c')
                    builder.append('2');
                else if(cleartext.charAt(i)>='d' && cleartext.charAt(i)<='f')
                    builder.append('3');
                else if(cleartext.charAt(i)>='g' && cleartext.charAt(i)<='i')
                    builder.append('4');
                else if(cleartext.charAt(i)>='j' && cleartext.charAt(i)<='l')
                    builder.append('5');
                else if(cleartext.charAt(i)>='m' && cleartext.charAt(i)<='o')
                    builder.append('6');
                else if(cleartext.charAt(i)>='p' && cleartext.charAt(i)<='s')
                    builder.append('7');
                else if(cleartext.charAt(i)>='t' && cleartext.charAt(i)<='v')
                    builder.append('8');
                else if(cleartext.charAt(i)>='w' && cleartext.charAt(i)<='z')
                    builder.append('9');
                
                //大写字母'A'~'Y'转换
                else if(cleartext.charAt(i)>='A' && cleartext.charAt(i)<='Y')
                    builder.append((char)(cleartext.charAt(i)+'a'-'A'+1));
                //大写字母'Z'单独处理
                else if(cleartext.charAt(i) == 'Z')
                    builder.append('a');
                
                //其他字符不做变换
                else
                    builder.append(cleartext.charAt(i));
            }
            System.out.println(builder.toString());
        }
    }
}