Question

我想知道是否还有另一种方法可以像我在这里那样不使用3个嵌套的for循环来解决此问题？我知道，如果在足够大的列表上测试该方法，则以这种方式嵌套循环很可能会引起很多问题。

这是问题：

from typing import List

def can_pay_with_three_coins(denoms: List[int], amount: int) -> bool:
    """Return True if and only if it is possible to form amount, which is a
    number of cents, using exactly three coins, which can be of any of the
    denominations in denoms.

    >>> can_pay_with_three_coins([1, 5, 10, 25], 36)
    True
    >>> can_pay_with_three_coins([1, 5, 10, 25], 37)
    False

    """

这是我的解决方案：

for i in range(len(denoms)):
    one_coin = denoms[i]
    for j in range(len(denoms)):
        another_coin = denoms[j]
        for k in range(len(denoms)):
            last_coin = denoms[k]
            if one_coin + another_coin + last_coin == amount:
                return True
return False

我确信还有另一种解决此问题的方法，只是我真的没有想到。
感谢您的帮助！

Answer 1

这是一个著名的问题，名为3 sum。

此解决方案的时间复杂度为O（n ^ 3），您可以实现O（n ^ 2）的算法，该算法在以下链接中以多种语言进行了解释和实现：

Find a triplet that sum to a given value

Answer 2

好吧，让我们用itertools作弊：）

import itertools
from typing import List


def can_pay_with_three_coins(denoms: List[int], amount: int) -> bool:
    """Return True if and only if it is possible to form amount, which is a
    number of cents, using exactly three coins, which can be of any of the
    denominations in denoms.

    >>> can_pay_with_three_coins([1, 5, 10, 25], 36)
    True
    >>> can_pay_with_three_coins([1, 5, 10, 25], 37)
    False

    """

    for variant in itertools.permutations(denoms, len(denoms)):
        if sum(variant[:3]) == amount:
            return True

    return False


print(can_pay_with_three_coins([1, 5, 10, 25], 36))
print(can_pay_with_three_coins([1, 5, 10, 25], 37))
print(can_pay_with_three_coins([1, 1, 5, 10, 25], 37))
print(can_pay_with_three_coins([1, 3, 5, 10, 25], 37))
print(can_pay_with_three_coins([20, 20, 20, 50], 60))

输出

True
False
False
False
True

三个嵌套的for循环会降低性能

2 个答案: