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问题描述
我想知道我的实施是否有效.我试图使用 python 找到该问题的最简单和低复杂度的解决方案.
I would like to know if my implementation is efficient. I have tried to find the simplest and low complex solution to that problem using python.
def count_gap(x):
"""
Perform Find the longest sequence of zeros between ones "gap" in binary representation of an integer
Parameters
----------
x : int
input integer value
Returns
----------
max_gap : int
the maximum gap length
"""
try:
# Convert int to binary
b = "{0:b}".format(x)
# Iterate from right to lift
# Start detecting gaps after fist "one"
for i,j in enumerate(b[::-1]):
if int(j) == 1:
max_gap = max([len(i) for i in b[::-1][i:].split('1') if i])
break
except ValueError:
print("Oops! no gap found")
max_gap = 0
return max_gap
让我知道你的意见.
推荐答案
您的实现将整数转换为基数为 2 的字符串，然后访问字符串中的每个字符.相反，您可以使用 << 和 & 访问整数中的每一位.这样做将避免访问每个位两次(首先将其转换为字符串，然后检查结果字符串中是否为1").它还将避免为字符串分配内存，然后为您检查的每个子字符串分配内存.
Your implementation converts the integer to a base two string then visits each character in the string. Instead, you could just visit each bit in the integer using << and &. Doing so will avoid visiting each bit twice (first to convert it to a string, then to check if if it's a "1" or not in the resulting string). It will also avoid allocating memory for the string and then for each substring you inspect.
您可以通过访问 1 << 来检查整数的每一位.0, 1 <<1, ..., 1 <<(x.bit_length).
You can inspect each bit of the integer by visiting 1 << 0, 1 << 1, ..., 1 << (x.bit_length).
例如:
def max_gap(x):
max_gap_length = 0
current_gap_length = 0
for i in range(x.bit_length()):
if x & (1 << i):
# Set, any gap is over.
if current_gap_length > max_gap_length:
max_gap_length = current_gap_length
current_gap_length = 0
else:
# Not set, the gap widens.
current_gap_length += 1
# Gap might end at the end.
if current_gap_length > max_gap_length:
max_gap_length = current_gap_length
return max_gap_length
