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

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