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Krovorgen
Asked:2021-10-24 15:01:03 +0000 UTC2021-10-24 15:01:03 +0000 UTC

帮助优化生产代码。Python

  • 772

找出列表中两个数字的最大乘积。数字也可以是负数。代码正在运行，但测试程序发出 Time Limit Exceeded。如何优化？是否可以将现有的 2 for 转换为列表理解是很有趣的，因为我没能在 3 小时内做到这一点。

def composition():
    s = [int(i) for i in input().split()]
    lists = []
    total = 0
    for i in range(len(s)):
        for j in range(len(s)):
            if s[i] * s[j] > total and i != j:
                total = s[i] * s[j]
                lists.append(s[i])
                lists.append(s[j])
    a = lists[-1]
    b = lists[-2]
    if a < b:
        return a, b
    return b, a
print(*composition())
python

3 个回答

  1. Best Answer

    不使用任何库的最短代码:)

    data = sorted([*map(int, input().split())])

print(data[0], data[1]) if data[0] * data[1] > data[-1] * data[-2] else print(data[-1], data[-2])

    更短的解决方案：

    data = sorted([*map(int, input().split())])

print(*max(data[:2], data[-2:], key=lambda obj: obj[0] * obj[1]))
    • 8

  2. 好吧，如果您完全正面决定，那只是一个lobeshnik，那么

    data = [*map(int, input().split())]

# оцениваем только положительные числа
max1pos = max(data[0], data[1])
max2pos = min(data[0], data[1])

for value in data:
    if value <= 0:
        continue

    if value >= max1pos:
        max1pos, max2pos = value, max1pos
    elif value > max2pos:
        max2pos = value

# оцениваем только отрицательные числа числа
max1neg = min(data[0], data[1])
max2neg = max(data[0], data[1])

for value in data:
    if value > 0:
        continue

    if value <= max1neg:
        max1neg, max2neg = value, max1neg
    elif value < max2neg:
        max2neg = value

# выводим результат
print(max1pos, max2pos) if max1pos * max2pos > max1neg * max2neg else print(max1neg, max2neg)

    更好（更短）：

    data = [*map(int, input().split())]

max1pos, max2pos = max(data[0], data[1]), min(data[0], data[1])
max1neg, max2neg = max2pos, max1pos

for value in data:
    if value <= 0:
        if value <= max1neg:
            max1neg, max2neg = value, max1neg
        elif value < max2neg:
            max2neg = value
    else:
        if value >= max1pos:
            max1pos, max2pos = value, max1pos
        elif value > max2pos:
            max2pos = value

# выводим результат
print(max1pos, max2pos) if max1pos * max2pos > max1neg * max2neg else print(max1neg, max2neg)
    • 3

  3. 我将假设序列中的数字是唯一的。Itertools.combinations非常适合查找所有可能的非重复组合。对于重复项 -  combination_with_replacement。

    from itertools import combinations

data = map(int, input().split()))
uq_pairs = combinations(data)   # только уникальные неповторяющиеся пары

result = max(
    uq_pairs, 
    key=lambda x: x[0] * x[1],  # функция-ключ для поиска максимума
)
    • 3

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