Codewars算法題（5）

第18題：(printer_error)

問題：

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.

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...

Sometimes there are problems: lack of colors, technical malfunction and a "bad" control string is produced e.g. aaaxbbbbyyhwawiwjjjwwm.

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.

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

#Examples:

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

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

問題描述：其實說簡單了就是找出一個字符串中不在‘abcdefghijklm’中的個數，並返回其所佔的比例

代碼實現：

def printer_error(s):
    c=0
    for i in range(0,len(s)):
        if s[i] not in 'abcdefghijklm':
            c+=1
    return str(c)+"/"+str(len(s))

評分較高代碼：

（1）

from re import sub
def printer_error(s):
    return "{}/{}".format(len(sub("[a-m]",'',s)),len(s)

在此簡單解釋一下sub()函數，sub()函數中第一個參數是要替換掉的字符，第二個參數是用什麼替換，第三個參數是

 替換哪個字符串的字符，該例中實現的是將s字符串中的[a-m]替換成空。具體使用方法可參見博客

format()函數實現格式化將format中的參數按前面”{}/{}”格式輸出。

（2）

def printer_error(s):
    return "%s/%s" % (len(s.translate(None, "abcdefghijklm")), len(s))

該方法中使用了translate（）函數，將s字符串中的“abcdefghijklm”刪除，與方法一異曲同工

（3）

def printer_error(s):
    import re
    return str(len(re.findall('[n-z]', s))) + "/" + str(len(s))

該方法用了findall()函數,方法詳解見博客

（4）

def printer_error(s):
    s1 = len([ x for x in s if ord(x) > ord('m')])
    return str(s1) +'/' + str(len(s))

第19題（切巧克力問題）

問題：

Description:

Your task is to split the chocolate bar of given dimension n x m into small squares. Each square is of size 1x1 and unbreakable. Implement a

function that will return minimum number of breaks needed.

For example if you are given a chocolate bar of size 2 x 1 you can split it to single squares in just one break, but for size 3 x 1 you must do two breaks.

If input data is invalid you should return 0 (as in no breaks are needed if we do not have any chocolate to split). Input will always be a non-negative integer.

代碼實現：

def breakChocolate(n, m):
    return ((n * m - 1) if n * m > 1 else 0)

評分較高代碼：

def breakChocolate(n, m):
    return max(n * m - 1, 0)

這一題很簡單不做過多解釋

201707261335 持續更新中~~~