Question

当前正在审查即将举行的考试，并且收到了此练习题，答案为8，但我不确定为什么。有人可以为我分解吗？我尝试跟踪它，但很快就令人困惑。

def confuse(s):     
   if len(s) <= 1:        
      return s     
   x = len(s) // 2     
   return confuse(s[:x]) + confuse(s[x:]) 

print(confuse('annoy'))

问题：不包括对confuse（'annoy'）的调用，在此函数终止之前进行了多少次递归调用？

谢谢！

Answer 1

您应该将此画成一棵树：

混淆（'annoy'）：

+- 'an' (half of annoy, rounded down)
|  +- 'a'
|  \- 'n'
\- 'noy'
   +- 'n'
   \- 'oy'
      +- 'o'
      \- 'y'

有八个电话。

Answer 2

它们被这样称呼：

confuse('annoy')
confuse('an') + confuse('noy')  # 5//2 = 2
confuse('a') + confuse('n') # 2//2 = 1
confuse('n') + confuse('oy') # 3//2 = 1
confuse('o') + confuse('y') # 2//2 = 1

因此，是8。

Answer 3

confuse('annoy') == confuse('an') + confuse('noy')
                 == confuse('a') + confuse('n') + confuse('n') + confuse('oy')
                 == 'a' + 'n' + 'n' + confuse('o') + confuse('y')
                 == 'a' + 'n' + 'n' + 'o' _ 'y'
                 == 'annoy'

计算在RHS上对confuse的呼叫，您将发现其中的8个。粗略地说，您会收到2**O(lg(n))个递归调用，它们的长度为n。

Answer 4

如果您将代码修改为

def confuse(s):
   print("called with -->", s)
   if len(s) <= 1:        
      return s     
   x = len(s) // 2     
   return confuse(s[:x]) + confuse(s[x:]) 

confuse('annoy')

您将获得

called with --> annoy
called with --> an
called with --> a
called with --> n
called with --> noy
called with --> n
called with --> oy
called with --> o
called with --> y

不包括对confuse（'annoy'）的调用，在此函数终止之前进行了多少次递归调用？ 9-1 = 8

Answer 5

我希望这段代码能帮助您理解

def confuse(s):     
   if len(s) <= 1:        
      return s     
   x = len(s) // 2
   print(s[:x], s[x:])
   return confuse(s[:x]) + confuse(s[x:]) 

print(confuse('annoy'))

Python-为什么要进行8次递归调用？

5 个答案: