我将如何编辑这个贪婪的函数来给我一个总和？

Question

所以，我想创建一个函数，它接受 int s 和数组 A，然后返回加起来为 s 的元素数组 A。如果没有子集，则应返回最接近 s 的值。

例如：

A = [12, 79, 99, 91, 81, 47]

s = 150

将返回：

[12, 91, 47]

为12 + 91 + 47是150。

以下是我到目前为止所拥有的。我究竟做错了什么？

def closest(s, A):

    if s == 0:

        return 0

    for i in range(len(A)):

        if A[i] <= s:

            return 1 + closest(s - A[i], A)

Answer 1

在您的情况下，代码是：import itertoolsdef itersum(nums, target):&nbsp; &nbsp; result = [seq for i in range(len(nums),0,-1) for seq in itertools.combinations(nums,i) if sum(seq) == target]&nbsp; &nbsp; if result != target:&nbsp; &nbsp; &nbsp; &nbsp;for j in range(target):&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;result1 = [seq for i in range(len(nums),0,-1) for seq in itertools.combinations(nums,i) if sum(seq) == target + j]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;result2 = [seq for i in range(len(nums),0,-1) for seq in itertools.combinations(nums,i) if sum(seq) == target - j]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;if (len(result1) + len(result2)) > 0:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;result = result1 if result1 > result2 else result2&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;break&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;&nbsp; &nbsp; return resultA = [12, 79, 99, 91, 81, 44]s = 150itersum(A, s)

Answer 2

该函数应该返回一个列表列表，因为可能有多个组合加起来为给定的总和：def closest(s, A):&nbsp; &nbsp; if s == 0:&nbsp; &nbsp; &nbsp; &nbsp; return [[]]&nbsp; &nbsp; o = []&nbsp; &nbsp; for i, n in enumerate(A):&nbsp; &nbsp; &nbsp; &nbsp; if n <= s:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; for c in closest(s - n, A[i + 1:]):&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; o.append([n] + c)&nbsp; &nbsp; return o以便：closest(150, [12, 79, 99, 91, 81, 47])返回：[[12, 91, 47]]然后：closest(10, [4, 5, 6, 2, 1, 3])返回：[[4, 5, 1], [4, 6], [4, 2, 1, 3], [5, 2, 3], [6, 1, 3]]