作为新手，我无法在程序中找到错误 [英] As a newbie I can't find the bug in the program

问题描述

我当时正在训练代码战卡塔，而卡塔是:

I was training a code wars kata and the kata was:

在工厂，打印机会打印盒子的标签.对于一种盒子，打印机必须使用颜色，为了简单起见，用从a到m的字母命名.

In a factory a printer prints labels for boxes. For one kind of boxes the printer has to use colors which, for the sake of simplicity, are named with letters from a to m.

打印机使用的颜色记录在控制字符串中.例如好"字样.控制字符串将为aaabbbbhaijjjm，这意味着打印机使用了三种颜色a，四种颜色b，一种颜色h然后一种颜色a ...

The colors used by the printer are recorded in a control string. For example a "good" control string would be aaabbbbhaijjjm meaning that the printer used three times color a, four times color b, one time color h then one time color a...

有时会出现问题:颜色不足，技术故障和坏"的现象.产生控制字符串，例如aaaxbbbbyyhwawiwjjjwwm的字母不是从a到m.

Sometimes there are problems: lack of colors, technical malfunction and a "bad" control string is produced e.g. aaaxbbbbyyhwawiwjjjwwm with letters not from a to m.

您必须编写一个函数printer_error，给定一个字符串将输出打印机的错误率作为一个字符串，该字符串表示一个有理数，其分子是错误数，而分母是控制字符串的长度.不要将此分数简化为更简单的表达式.

You have to write a function printer_error which given a string will output the error rate of the printer as a string representing a rational whose numerator is the number of errors and the denominator the length of the control string. Don't reduce this fraction to a simpler expression.

该字符串的长度大于或等于1，并且仅包含从a到z的字母.

The string has a length greater or equal to one and contains only letters from a to z.

s="aaabbbbhaijjjm"
error_printer(s) => "0/14"

s="aaaxbbbbyyhwawiwjjjwwm"
error_printer(s) => "8/22"

作为一个新手，我试图尝试它.我的程序是这样的:

and as a newbie, I tried to attempt it . My program is like this:

public class Printer {
    
    public static String printerError(String s) {
      int printErr = 0;
      char end = 110;
      int i = 0;
        while (i < s.length()){
          if(s.charAt(i) > end ){
          printErr++;
          }
          i++;
        }
        String rate = String.format("%d/%d",printErr , s.length());
        return rate;
    }
}

它通过了测试，但是在提交Kata时，计数器缺少1或2个数字.有人可以帮忙吗?

It passed the test but while submitting the Kata the counter was missing 1 or 2 numbers. Can anyone help?

推荐答案

实际上，您可以仅使用<和>来检查字符是否在Java中.您的逻辑是正确的-但由于您是新手"，因此已使用while循环重新创建了for循环的功能.无需这样做-这就是我们有for循环的原因.

You can actually just use < and > to check if a character is in some range in java. Your logic is sound - but since you are a "newbie", you have re-created the functionality of a for-loop with your while loop. No need to do this - that's why we have for-loops.

请参阅下面的调整方法:

See the adjusted method below:

    public String printerError(String s) {
        int printErr = 0;

        for (int i = 0; i < s.length(); i++) {
            // assuming the input rules hold true, we really only need the second condition
            if (s.charAt(i) < 'a' || s.charAt(i) > 'm') {
                printErr++;
            }
        }

        return String.format("%d/%d", printErr, s.length());
    }

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