问题描述

输入

sum_possible(2017, [4, 2, 10]) # -> False

使用any导致解决方案挂起/花费较长时间

def sum_possible(amount, numbers, cache = None):
  if cache is None:
    cache = {}
  if amount in cache:
    return cache[amount]
  if amount == 0:
    return True
  if amount < 0:
    return False
  cache[amount] = any([sum_possible(amount - number, numbers, cache) for number in numbers])
  return cache[amount]

使用for循环在合理的时间内解决问题

def sum_possible(amount, numbers, cache = None):
  if cache is None:
    cache = {}
  if amount in cache:
    return cache[amount]
  if amount == 0:
    return True
  if amount < 0:
    return False
  
  for number in numbers:
    if sum_possible(amount - number, numbers, cache):
      cache[amount] = True
      return True
  
  cache[amount] = False
  return False

我以为any会短路吗？如果它遇到True？

，则有效地提前return True

---

解决方案

any()会短路，但您正在构建一个列表以首先传递给any。

cache[amount] = any([sum_possible(amount - number, numbers, cache) for number in numbers])

首先评估列表理解--并且它是迫切的：

[sum_possible(amount - number, numbers, cache) for number in numbers]

用生成器表达式替换它-这应该可以用(惰性求值)：

cache[amount] = any(sum_possible(amount - number, numbers, cache) for number in numbers)