递归函数无法正常工作

发布于 2024-09-04 09:13:17 字数 577 浏览 3 评论 0原文

我很难弄清楚这里出了什么问题：

class iterate():
    def __init__(self):
        self.length=1
    def iterated(self, n):
        if n==1:
            return self.length
        elif n%2==0:
            self.length+=1
            self.iterated(n/2)
        elif n!=1:
            self.length+=1
            self.iterated(3*n+1)

例如，

x=iterate()
x.iterated(5)

输出 None。它应该输出 6，因为长度如下所示： 5 --> 16 --> 8 --> 4 --> 2 --> 1

经过一些调试，我发现 self.length 正确返回，但递归中出现问题。我不太确定。感谢您的任何帮助。

原文

I'm having quite a hard time figuring out what's going wrong here:

class iterate():
    def __init__(self):
        self.length=1
    def iterated(self, n):
        if n==1:
            return self.length
        elif n%2==0:
            self.length+=1
            self.iterated(n/2)
        elif n!=1:
            self.length+=1
            self.iterated(3*n+1)

For example,

x=iterate()
x.iterated(5)

outputs None. It should output 6 because the length would look like this:
5 --> 16 --> 8 --> 4 --> 2 --> 1

After doing some debugging, I see that the self.length is returned properly but something goes wrong in the recursion. I'm not really sure. Thanks for any help.

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情仇皆在手 2024-09-11 09:13:18

在两个 elif 块中，进行递归调用后不会返回值。在递归调用iterated 之前，您需要一个return（例如return self.iterated(n/2)）。如果您没有显式return，该函数将返回None。

这将解决这个问题，但有一种方法可以使您的代码更简单：您实际上不需要成员length。相反，您可以将递归调用的结果加 1：

def iterated(n):
    if n==1:
        return 1
    elif n%2==0:
        return 1 + iterated(n/2)
    else:
        return 1 + iterated(3*n+1)

print(iterated(5))

这不需要位于类中，因为不需要任何成员。

In the two elif blocks, you don't return a value after making the recursive call. You need a return before the recursive calls to iterated (e.g. return self.iterated(n/2)). If you don't explicitly return, the function will return None.

That will fix this issue, but there is a way to make your code simpler: You don't actually need the member length. Instead, you can add 1 to the result of the recursive call:

def iterated(n):
    if n==1:
        return 1
    elif n%2==0:
        return 1 + iterated(n/2)
    else:
        return 1 + iterated(3*n+1)

print(iterated(5))

This doesn't need to be in a class, since there is no need for any members.

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与酒说心事 2024-09-11 09:13:18

您缺少返回语句：

class iterate():
    def init(self):
        self.length=1
    def iterated(self, n):
        if n==1:
            return self.length
        elif n%2==0:
            self.length+=1
            **return** self.iterated(n/2)
        elif n!=1:
            self.length+=1
            **return** self.iterated(3*n+1)

You're missing the return statements:

class iterate():
    def init(self):
        self.length=1
    def iterated(self, n):
        if n==1:
            return self.length
        elif n%2==0:
            self.length+=1
            **return** self.iterated(n/2)
        elif n!=1:
            self.length+=1
            **return** self.iterated(3*n+1)

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