我尝试在 HackerRank 上解决一个名为“Apple and orange”的问题，代码如下：

def countApplesAndOranges(s, t, a, b, apples, oranges):
    count_apples = 0
    count_oranges = 0
    x = [x for x in range(s, t 1)]
    pos_apple = [apple   a for apple in apples]
    pos_orange = [orange   b for orange in oranges]
    for i in x:
        for j in pos_apple:
            if j == i:
                count_apples  =1
        for l in pos_orange:
            if l == i:
                count_oranges  = 1
    print(count_apples)
    print(count_oranges)

该代码有效。但是，当我尝试提交它时，它通过了前 3 个测验，其余测验失败，但出现例外“因超时而终止”。我检查了其中一项测验的输入，它需要处理大量资料，您可以在此处查看资料：https : //hr-testcases-us-east-1.s3.amazonaws.com/ 25220/input03.txt?AWSAccessKeyId=AKIAR6O7GJNX5DNFO3PV&Expires=1642016820&Signature=J4ypdP0YzRxcOWp+y5XaD5ITeMw=&response-content-type=text/plain

它失败了，因为通过我的 IDE 处理具有相同输入的代码需要大约 2 分钟，但 HackerRank 测验限制为 10 秒。

我需要你的帮助来优化代码并让它运行得更快。

嵌套回圈似乎是这里最大的问题，但我不知道应该用什么替换。

uj5u.com热心网友回复：

检查对每个每个苹果/橙坐标范围内将你的代码的运行时复杂到O(n * a n * o)哪里n是家之长，a是苹果的数量，o是橘子的数量。理想情况下，您的代码必须在O(a o).

这是您的解决方案的重构版本：

def countApplesAndOranges(s, t, a, b, apples, oranges):
    count_apples = 0
    count_oranges = 0
    for apple in pos_apple:
        if s <= apple   a <= t:
            count_apples  =1
    for orange in pos_orange:
        if s <= orange   b <= t:
            count_oranges  = 1
    print(count_apples)
    print(count_oranges)

uj5u.com热心网友回复：

如果可以，请不要构建中间串列，如果确实需要这样做（这不是问题的情况），请改用生成器

并且尽量不要重复自己，尽可能用函式分解

def countApplesAndOranges(home_start, home_end, tree_apple, tree_orange, apples, oranges):
    def count_fruits(home_start, home_end, tree, fruits): 
        count = 0
        for fruit in fruits:
            if home_start <= tree   fruit <= home_end:
                count  =1
        return count
    
    print(count_fruits(home_start, home_end, tree_apple, apples))
    print(count_fruits(home_start, home_end, tree_orange, oranges))

uj5u.com热心网友回复：

我想我会sum()以几个串列推导作为起点：

apple_hits = sum(
    1 for apple in apples
    if house_min <= apple   apple_tree_origin <= house_max
)

不过，向我指出的一件事是，从该测验的两侧减去 apple_tree_origin 不应该改变它：

apple_hits = sum(
    1 for apple in apples
    if house_min - apple_tree_origin <= apple <= house_max - apple_tree_origin
)

现在我们可能会观察到house_min - apple_tree_origin可以预先计算，剩下的我的答案是：

def countApplesAndOranges1(s, t, a, b, apples, oranges):
    house_min = s
    house_max = t

    apple_tree_origin = a
    orange_tree_origin = b

    house_min_apples = house_min - apple_tree_origin
    house_max_apples = house_max - apple_tree_origin

    house_min_oranges = house_min - orange_tree_origin
    house_max_oranges = house_max - orange_tree_origin

    apple_hits = sum(1 for apple in apples if house_min_apples <= apple <= house_max_apples)
    orange_hits = sum(1 for orange in oranges if house_min_oranges <= orange <= house_max_oranges)
    return apple_hits, orange_hits

有了你提供的测验资料，我得到了(18409, 19582)。希望这是正确的。

随意timeit反对其他解决方案：

import timeit

setup = """
with open("apples_oranges.txt") as file_in:
    s,t = list(map(int, file_in.readline().split()))
    a,b = list(map(int, file_in.readline().split()))
    m,n = list(map(int, file_in.readline().split()))
    apples = list(map(int, file_in.readline().split()))
    oranges = list(map(int, file_in.readline().split()))

def countApplesAndOranges_jonsg(s, t, a, b, apples, oranges):
    house_min = s
    house_max = t

    apple_tree_origin = a
    orange_tree_origin = b

    house_min_apples = house_min - apple_tree_origin
    house_max_apples = house_max - apple_tree_origin

    house_min_oranges = house_min - orange_tree_origin
    house_max_oranges = house_max - orange_tree_origin

    apple_hits = sum(1 for apple in apples if house_min_apples <= apple <= house_max_apples)
    orange_hits = sum(1 for orange in oranges if house_min_oranges <= orange <= house_max_oranges)
    return apple_hits, orange_hits
"""

print(timeit.timeit("countApplesAndOranges_jonsg(s, t, a, b, apples, oranges)", setup=setup, number=1000))

uj5u.com热心网友回复：

感谢你们！

我想通了，并使用了

if s <= apple   a <= t:

在我的代码中，因为它是摆脱嵌套回圈的最简单的解决方案。它作业得很好，让我想知道我怎么没有想到它。

我真的很喜欢您的所有解决方案，并感谢您的帮助！