递归方法countAlphabet [英] recursive method countAlphabet

问题描述

如何接收字符串作为其参数并返回字符串中的字母数.

例如:
字符串:价格为$ 70."
字母数:10

这是我的代码，但是如何仅计算字母?

how to receive a String as its parameter and return the number of letters count in the String.

example:
String: "The price is $70."
Alphabets count: 10

this my code, but how to count only alphabets?

public class MyRecursion
{
    //driver method
    public static void main(String[] args)
    {
        //Example of a string to test with isPalindrome
        String str1 = "Malam";

        if( isPalindrome(str1))
        System.out.println("Word: " + str1 + " is a Palindrome");
        else
        System.out.println("Word: " + str1 + " is NOT a Palindrome");
    }//end main

    //isPalindrome method
    public static boolean isPalindrome(String str)
    {
        if (str.length() <= 1)
        return true;
        else
        {
            str = str.toLowerCase();
            char first = str.charAt(0);
            char last = str.charAt(str.length()-1);
            if (first == last )
            return isPalindrome(str.substring(1, str.length()-1));
            else
            return false;
        }
    }//end isPalindrome method
}//end class

推荐答案

70."
字母数:10

这是我的代码，但是如何仅计算字母?

70."
Alphabets count: 10

this my code, but how to count only alphabets?

public class MyRecursion
{
    //driver method
    public static void main(String[] args)
    {
        //Example of a string to test with isPalindrome
        String str1 = "Malam";

        if( isPalindrome(str1))
        System.out.println("Word: " + str1 + " is a Palindrome");
        else
        System.out.println("Word: " + str1 + " is NOT a Palindrome");
    }//end main

    //isPalindrome method
    public static boolean isPalindrome(String str)
    {
        if (str.length() <= 1)
        return true;
        else
        {
            str = str.toLowerCase();
            char first = str.charAt(0);
            char last = str.charAt(str.length()-1);
            if (first == last )
            return isPalindrome(str.substring(1, str.length()-1));
            else
            return false;
        }
    }//end isPalindrome method
}//end class

我'和CPallini在一起-这里有很强的作业感.所以，没有代码！

我不会进行递归-这里没有需要递归的内容.只需循环遍历字符串中的每个字符，并且-如果它是字母，则将其添加到计数中.

简单！

I''m with CPallini - there is a strong sense of homework here. So, no code!

I wouldn''t make this recursive - there is nothing here that requires recursion. Just loop through each of the characters in the string and - if it is alphabetic - add one to a count.

Simples!

示例递归 [ ANSI可打印字符 [ ^ ].选中"Dec"此表上的(十进制)列以供参考.

Recursion explained with examples[^]

The trick is to check the character if it is a int or char or something else.

You can accomplish the given task by trying to convert the character into a int - a NumberFormatException will appear if it is not a int. I think this is what your teacher has in mind.

Other solution (more elegant): convert the character into a int and check if it is within the range (ANSI Printable Characters[^] of a letter. Check the "Dec" (decimal) column on this table for reference.

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