Question

我想出了与解决方案中提到的自底向上的动态编程方法相同的方法：

import math


class Solution:
    def coinChange(self, coins, amount):
        fewest = [0] * (amount + 1)
        for i in range(1, amount + 1):
            fewest[i] = 1 + min(
                (fewest[i - coin] for coin in
                    [c for c in coins if i - c >= 0]),
                default=math.inf)
        return fewest[amount] if fewest[amount] < math.inf else -1

这是我用来验证解决方案的一些pytest测试用例：

def test_1():
    assert Solution().coinChange([1, 2, 5], 1) == 1


def test_2():
    assert Solution().coinChange([1, 2, 5], 2) == 1


def test_3():
    assert Solution().coinChange([1, 2, 5], 3) == 2


def test_4():
    assert Solution().coinChange([1, 2, 5], 4) == 2


def test_5():
    assert Solution().coinChange([1, 2, 5], 5) == 1


def test_67():
    assert Solution().coinChange([1, 2, 5], 6) == 2
    assert Solution().coinChange([1, 2, 5], 7) == 2


def test_8():
    assert Solution().coinChange([1, 2, 5], 8) == 3


def test_11():
    assert Solution().coinChange([1, 2, 5], 11) == 3


def test_no_way():
    assert Solution().coinChange([2], 3) == -1

问题是我收到“超过时间限制”错误：

但是，如果我复制此测试用例并在本地运行，我发现该算法仅需0.02s：

import pytest

def test_time_limit_exceeded():
    Solution().coinChange(
        [205, 37, 253, 463, 441, 129, 156, 429, 101, 423, 311],
        6653)


if __name__ == "__main__":
    pytest.main([__file__, '--duration', '1'])

导致以下输出：

============================= test session starts ==============================
platform darwin -- Python 3.6.6, pytest-3.8.1, py-1.6.0, pluggy-0.7.1
rootdir: /Users/kurtpeek/GoogleDrive/CodeRust, inifile:
collected 11 items

coin_changing_leetcode2.py ...........                                   [100%]

=========================== slowest 1 test durations ===========================
0.02s call     coin_changing_leetcode2.py::test_time_limit_exceeded
========================== 11 passed in 0.07 seconds ===========================

有任何想法为什么LeetCode未能通过该实现？

Answer 1

似乎这件作品：

  fewest[i] = 1 + min(
            (fewest[i - coin] for coin in
                [c for c in coins if i - c >= 0]),
            default=math.inf)

检查所有硬币，过滤适当的硬币。

但是您可以对普通名词进行排序，并且只对给定的i遍历足够小的名词。

Answer 2

我意识到对于每个[c for c in coins if i - c >= 0]，列表理解len(coins)被评估i次，而对于每个i，它只需要评估一次。接受了这种稍微重构的解决方案：

import math


class Solution:
    def coinChange(self, coins, amount):
        fewest = [0] * (amount + 1)
        for i in range(1, amount + 1):
            eligible_coins = [c for c in coins if i - c >= 0]
            fewest[i] = 1 + min(
                (fewest[i - coin] for coin in eligible_coins),
                default=math.inf)
        return fewest[amount] if fewest[amount] < math.inf else -1

尽管如此，它仍然是针对该问题的Python 3解决方案中排在最后10％的位置之一

LeetCode的硬币找零问题超出时间限制

2 个答案: