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我需要使用递归为 collat​​z 猜想编写 python 代码，其中提示用户输入一个正整数，如果偶数乘以 3，则该数字除以 2，如果奇数则加 1，并且此序列一直持续到值等于一。我还必须提示用户选择如何显示序列，无论是标准的计算方式、反转方式还是回文方式（向前和向后，即 86324895159842368） 以下是我目前所拥有的。我计算序列本身没有问题，但我不确定如何实现第二个参数。每当我尝试将方向定义为 F、B 或 P 时，都会出现一堆错误。在我需要去的地方的任何帮助将不胜感激

## CollatzRecursion


###
# define your RECURSIVE version of Collatz below this comment so that it runs
# correctly when called from below.

def Collatz(m):
    seq = [m]
    if m < 1:
       return []
    while m > 1:
       if m % 2 == 0:
         m = m / 2
       else:
         m = 3 * m + 1 
       seq.append(m)    
       if displaymode (F) :
         return seq
       if displaymode  (B) :
         seq = seq[::-1]
         return seq
       if displaymode (P) :


#
# REQUIREMENTS:
#  a) The correct sequence must be printed, all the values on one line.
#  b) Your Collatz function must use recursion.
#  c) Aside from two arguments accepting a positive integer value and the letter
#     F, B, or P; your Collatz function MAY NOT use any other internal variables.
#  d) Your Collatz function may accept only the two arguments described in (c).
#  e) If the second argument is 'F', the sequence should b printed in its
#       naturally generated order.
#     If the second argument is 'B', the sequence should be printed in reverse.
#     If the second argument is 'P', then a palindrome of the sequence values should 
#       be printed (see http://en.wikipedia.org/wiki/Palindrome).  In this case
#       it doesn't matter if your function prints the first value as 1 or the
#       value provided by the user.
###







###
# Do NOT alter Python code below this line
###
m = input( "Enter a positive integer value: " )
displaymode = ''  # initialize to anything not F, B, P
while displaymode not in ['F', 'B', 'P'] :
    displaymode = raw_input( "Choose a display mode:  F=forward, B=backward,         P=palindrome: " )

Collatz( m, displaymode )
print
4

1 回答 1

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首先，您当前的方法不像请求的那样递归。

我个人会做这样的事情（符合递归的需要和缺少额外的变量）：

def collatz(acc, mode):
    if isinstance(acc, int):
        acc = [acc]
    if acc[-1] == 1:
        if mode not in 'FBP':
            raise Exception('Unsupported display type')
        if mode == 'F':
            print ''.join(map(str, acc))
        elif mode == 'B':
            print ''.join(map(str, acc[::-1]))
        else:
            print ''.join(map(str, acc + acc[::-1]))
        return
    collatz(acc + [acc[-1] * 3 + 1 if acc[-1] % 2 else acc[-1] / 2], mode)

示例用法：

>>> collatz(11, 'F')
1134175226134020105168421
>>> collatz(11, 'B')
1248165102040132652173411
>>> collatz(11, 'P')
11341752261340201051684211248165102040132652173411
于 2014-04-23T19:14:08.827 回答