Question

我有一系列长篇,从中得出以下顺序:

b[0] = 0
b[1] = b[0] + a[0]
b[2] = b[1] + a[0]
b[3] = b[2] + a[1]
b[4] = b[3] + a[0]
b[5] = b[4] + a[1]
b[6] = b[5] + a[2]
b[7] = b[6] + a[0]
b[8] = b[7] + a[1]
b[9] = b[8] + a[2]
b[10] = b[9] + a[3]
#etc.

a 可能含有不积极值。我需要找到最大部分。我只收到O(n^2)的解决办法。是否有更快的办法?

def find_max(a):
  b = [0]
  i = 0
  count = 0
  while i < len(a):
    j = 0
    while j <= i:
      b.append(b[count] + a[j])
      count += 1
      j += 1
    i += 1
  return max(b)

Answer 1

仍然精细,但似乎较快五倍,我发现它更清楚/更明确了<>?

from itertools import chain, accumulate

def find_max_2(a):
    def prefixes():
        prefix = []
        for x in a:
            prefix.append(x)
            yield prefix
    increments = chain.from_iterable(prefixes())
    return max(accumulate(increments, initial=0))

独角兽器似乎仍比你更快四倍:

from itertools import chain, islice, repeat, accumulate

def find_max_fun(a):
    return max(accumulate(chain(*map(islice, repeat(a), range(1, len(a) + 1))), initial=0))

Or with a :

def Shlemiel(xs):
    path = []
    for x in xs:
        path.append(x)
        yield from path

def find_max_fun(a):
    return max(accumulate(Shlemiel (a), initial=0))

*** 虽然这或许只是因为你使用了相当不成熟的代码。我可以这样写:

def find_max(a):
    b = [0]
    for i in range(1, len(a) + 1):
        for x in a[:i]:
            b.append(b[-1] + x)
    return max(b)

Answer 2

这里是O(n)解决办法:

def find_max(a):
  b = [0]
  i = 1
  max_a = 0

  while i <= len(a):
    for j in range(0, max_a + 1):
      b.append(b[i-1] + a[j])
      i += 1
      if i >= len(a):
        break
    max_a += 1
  return max(b)

测试:

print(find_max([1, 2, 3, -4, -2, 1])) # 7

由于产出减少,你可以看到进展:

def find_max(a, debug=False):
  b = [0]
  if debug:
    print (f b[0] = {b[0]} )
  i = 1
  max_a = 0

  while i <= len(a):
    for j in range(0, max_a + 1):
      b.append(b[i-1] + a[j])
      if debug:
        print (f b[{i}] = b[{i-1}] + a[{j}] )
      i += 1
      if i >= len(a):
        break
    max_a += 1
  return max(b)

<代码>(定义_max([1、2、3、4、2、1]、debug=True])

b[0] = 0
b[1] = b[0] + a[0]
b[2] = b[1] + a[0]
b[3] = b[2] + a[1]
b[4] = b[3] + a[0]
b[5] = b[4] + a[1]
7

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