Python中断/返回子集总和的递归函数

Question

我正在尝试实现this function来确定任何子集的总和是否等于目标值：

def subset_sum(numbers, target, partial=[]):
    s = sum(partial)

    # check if the partial sum is equals to target
    if s == target: 
        print "sum(%s)=%s" % (partial, target)
    if s >= target:
        return  # if we reach the number why bother to continue

    for i in range(len(numbers)):
        n = numbers[i]
        remaining = numbers[i+1:]
        subset_sum(remaining, target, partial + [n]) 

但是，我希望一旦遇到目标实例，该函数就可以“中断” /返回，这样它就不会继续寻找更多对象。

我已经尝试过了，但是仍然继续：

def subset_sum(numbers, target, partial=[]):
    s = sum(partial)

    # check if the partial sum is equals to target
    if s == target: 
        print "sum(%s)=%s" % (partial, target)
        return   // should stop once reaches one instance
    if s >= target:
        return  # if we reach the number why bother to continue

    for i in range(len(numbers)):
        n = numbers[i]
        remaining = numbers[i+1:]
        subset_sum(remaining, target, partial + [n]) 

Answer 1

使subset_sum返回布尔值：是否找到该子集。检查递归调用的结果。

def subset_sum(numbers, target, partial=[]):
    s = sum(partial)

    # check if the partial sum is equals to target
    if s == target: 
        print "sum(%s)=%s" % (partial, target)
        return True   # should stop once reaches one instance
    if s > target:
        return False # if we reach the number why bother to continue

    for i in range(len(numbers)):
        n = numbers[i]
        remaining = numbers[i+1:]
        if subset_sum(remaining, target, partial + [n]):
            return True
    return False

更好的方法可能是返回答案（即partial或None。