如何将带有数字和字母的字符串分成数字？

Question

public static String carRentalCode(String licensePlate) {
    // precondition:    licensePlate is a valid plate as described in pdf
    // postcondition:   return the car rental code for the licensePlate as described
    int values = 0;
    int ascii = 0;
    int sum = 0;
    int convert = 0;
    int count = 0;
    int a = 0;
    char letter;
    String license;
    char[] ch = new char[licensePlate.length()];
    for (int i = 0; i < licensePlate.length(); i++) { 
        ch[i] = licensePlate.charAt(i); 
    }
    for(int i = 0; i< ch.length; i++)
    {
        if(Character.isDigit(ch[i]))
        {
            values += Character.getNumericValue(ch[i]);
        }
        if(Character.isAlphabetic(ch[i]))
        {
            ascii += (int) ch[i];
            count++;
        }
    }
    sum = values + ascii;
    convert = sum%26 + 65;
    letter = (char)convert;
    String[] letters = licensePlate.split("\\d+");
    String lowercase = Arrays.toString(letters).toLowerCase();
    String code = Integer.toString(sum);
    return letter + code + lowercase;

Here I am supposed to return a car rental code based on a given license plate (the tester plate is "123ABC456" ). The only problem I have left is that my lowercase returns as [[,abc]] as opposed to the correct answer which is [abc] .

How do you fix the empty space in front, and there being two pairs of brackets instead of one? Or is there any other way to only obtain the letters of a string (ABC) , separated from the digits (123 and 456) and return that instead?

Expected: L219[abc]

Actual: L219[[, abc]]

Answer 1

I changed some of your code, use String.toCharArray() to create an array of characters instead of copying it char by char, and replace all your digits by empty string to get only the alphabetic characters

public static String carRentalCode(String licensePlate) {
    // precondition:    licensePlate is a valid plate as described in pdf
    // postcondition:   return the car rental code for the licensePlate as described
    int values = 0;
    int ascii = 0;
    int sum = 0;
    int convert = 0;
    char letter;

    char[] ch = licensePlate.toCharArray();

    for (int i = 0; i < ch.length; i++) {
        if (Character.isDigit(ch[i])) {
            values += Character.getNumericValue(ch[i]);
        }
        if (Character.isAlphabetic(ch[i])) {
            ascii += ch[i];
        }
    }
    sum = values + ascii;
    convert = sum % 26 + 65;
    letter = (char) convert;

    String lowercase = licensePlate.replaceAll("\\d+", "").trim().toLowerCase();
    String code = Integer.toString(sum);
    return letter + code + "[" + lowercase + "]";
}

Answer 2

I just wrote the following code which extracts a word (without a number) from a string:

import java.util.*;
import java.lang.*;
import java.io.*;

import java.util.regex.Matcher; 
import java.util.regex.Pattern; 

class Ideone
{
    static void extractWord(String str) 
    { 
        String regex = "[a-zA-Z]+"; 

        // compiling regex 
        Pattern p = Pattern.compile(regex); 

        // Matcher object 
        Matcher m = p.matcher(str); 

        while(m.find()) 
        { 
            System.out.println(m.group());
        } 
    } 

    public static void main (String[] args) 
    { 
        String str = "123ABC456"; 

        extractWord(str); 
    }
}

Hope this helps you.

Answer 3

Just as an example of a more functional approach:

String lowercase = licensePlate.chars()
                .filter(Character::isAlphabetic)
                .map(Character::toLowerCase)
                .mapToObj(c -> Character.toString((char) c))
                .reduce((a, b) -> a + b)
                .orElse("");

Answer 4

刚回来怎么样

letter + code + "[" + letters[1].toLowerCase() + "]"