子数组的元素总和（可以重复）等于Python中的目标值

Question

def howSum(targetSum, numbers):
    if targetSum == 0:
        return []
    if targetSum < 0:
        return None
    for num in numbers:
        remainder = targetSum - num
        remainderResult = howSum(remainder, numbers)
        if remainderResult != None:
            return [remainderResult, num]
    return None

我已尝试此代码并得到答案 [[[[[], 2], 2], 2], 2] 而不是 [2, 2, 2, 2]。我对编程和堆栈交换很陌生。请帮帮我。提前致谢

Answer 1

使用序列解包。

代替：

            return [remainderResult, num]

用途：

            return [*remainderResult, num]

完整代码：

def howSum(targetSum, numbers):
    if targetSum == 0:
        return []
    if targetSum < 0:
        return None
    for num in numbers:
        remainder = targetSum - num
        remainderResult = howSum(remainder, numbers)
        if remainderResult != None:
            return [*remainderResult, num]
    return None'

输出：

[2, 2, 2, 2]