[英]Python tail recursion optimization in this algo?
我遇到了以下问题:让我们考虑以下编码表:
0 -> A
1 -> B
2 -> C
...
"0" -> 1 (word is A)
"10" -> 2 (words are BA and K)
"121" -> 3 (words are BCB, MB and BV)
我写了这个算法
import string
sys.setrecursionlimit(100) # limit the number of recursions to 100
# {0: 'a', 1: 'b', etc.}
dict_values = dict(zip(range(26), string.ascii_lowercase))
def count_split(list_numbers: list) -> int:
if len(list_numbers) == 0:
return 0
elif len(list_numbers) == 1:
return 1
elif len(list_numbers) == 2:
if int(list_numbers[0]) == 0:
return 1
elif 10 <= int("".join(list_numbers)) <= 25:
return 2
else:
return 1
else: # list_numbers contains at least three elements
if count_split(list_numbers[:2]) == 2:
return count_split(list_numbers[1:]) + count_split(list_numbers[2:])
else:
return count_split(list_numbers[1:])
def get_nb_words(code: str) -> int:
return count_split(list(code))
# print(get_nb_words("0124")) -> 3
# print(get_nb_words("124")) -> 3
# print(get_nb_words("368")) -> 1
# print(get_nb_words("322")) -> 2
# print(get_nb_words("12121212121212121212")) -> 10946
令人惊讶的是,该算法适用于最后一个示例"12121212121212121212"
。 我预计会超过递归次数,因为在每个步骤中,函数count_split
在列表中被调用两次。 这样一来,调用次数就远远超过100次(甚至超过1000次)!
与此同时,我在 stackoverflow 上发现这篇文章说尾递归没有在 Python 中优化,所以我有点惊讶!
有人可以向我解释为什么在这个算法中没有超过递归限制吗?
您关心递归深度,即调用堆栈的最大深度(高度?)。
这是一种经验方法:(测量深度的代码)
import string, sys
sys.setrecursionlimit(100) # limit the number of recursions to 100
dict_values = dict(zip(range(26), string.ascii_lowercase))
stack = []
max_depth = 0
def count_split(list_numbers: list) -> int:
global max_depth
stack.append(None)
max_depth = max(max_depth, len(stack))
if len(list_numbers) == 0:
return 0
elif len(list_numbers) == 1:
return 1
elif len(list_numbers) == 2:
if int(list_numbers[0]) == 0:
return 1
elif 10 <= int("".join(list_numbers)) <= 25:
return 2
else:
return 1
else: # list_numbers contains at least three elements
if count_split(list_numbers[:2]) == 2:
stack.pop()
result = count_split(list_numbers[1:]) + count_split(list_numbers[2:])
stack.pop(); stack.pop()
return result
else:
result = count_split(list_numbers[1:])
stack.pop()
return result
def get_nb_words(code: str) -> int:
return count_split(list(code))
print(get_nb_words("12121212121212121212"))
print(max_depth) # 20
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