通过递归只获取数字中的奇数位 [英] getting only the odd digits in a number with recursion

问题描述

所以我的问题是我有一个像 123 这样的数字，并且根据倾斜度，我希望结果是 13.

问题是，首先使用 im 的方法将得到取反结果(例如 31)，其次我在末尾得到一个不应该存在的零，而不是将数字相加，而不是将它们和我不明白为什么.顺便说一句，我不能使用字符串

澄清一下:

我的输出:

<预><代码>>>>apenas_digitos_impares(123)40

正确的输出:

<预><代码>>>>apenas_digitos_impares(123)13

程序:

def apenas_digitos_impares(n):如果 n == 0:返回 0elif (n%10)%2 == 0:return apenas_digitos_impares(n//10)elif (n%10)%2 == 1:返回 10*(n%10) + apenas_digitos_impares(n//10)

解决方案

你可以这样做 -

def apenas_digitos_impares(n):如果 n == 0:返回 0elif (n%10)%2 == 0:return apenas_digitos_impares(n//10)elif (n%10)%2 == 1:# 包括数字并递归剩余...返回 (n%10) + 10*apenas_digitos_impares(n//10)打印(apenas_digitos_impares(123))

输出:

13

您的代码所需的唯一更改是在函数的最后一行.

• 您将只包括奇数(由 n%10 完成)和，

• 继续(或递归)检查剩余的数字.你需要将下一个数字乘以 10，所以 - 10*apenas_digitos_impares(n//10)

So my problem is this i have an number like 123 and as the tilte sugests i want the result to be 13.

The problem is that firstly with the method im using im going to get the invert result (31 for example), secondly im getting a zero at the end that shouldn't be there and instead of joining the digits its summing them and i dont understand why. BTW i cant use strings

So to clarify:

My output:

>>> apenas_digitos_impares(123)
40

Correct output:

>>> apenas_digitos_impares(123)
13

program:

def apenas_digitos_impares(n):
    if n == 0:
        return 0
    elif (n%10)%2 == 0:
        return apenas_digitos_impares(n//10)
    elif (n%10)%2 == 1:
        return 10*(n%10) + apenas_digitos_impares(n//10)

解决方案

You could do it as follows -

def apenas_digitos_impares(n):
    if n == 0:
        return 0
    elif (n%10)%2 == 0:
        return apenas_digitos_impares(n//10)
    elif (n%10)%2 == 1:
        # Include the digit and recurse for remaining...
        return (n%10) + 10*apenas_digitos_impares(n//10)
        
print(apenas_digitos_impares(123))

OUTPUT :

13

The only change that your code needed was in the last line of the function.

• You will just include the odd digit(done by n%10) and,

• move on(or recurse) to check for remaining digits. You need to multiply next digit by 10, so - 10*apenas_digitos_impares(n//10)

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