Question

我希望按特定顺序枚举解决方案。目前，使用以下代码：

def balls_in_baskets(balls=1, baskets=1):
    if baskets == 1:
        yield [balls]
    elif balls == 0:
        yield [0]*baskets
    else:
        for i in range(balls+1):
            for j in balls_in_baskets(balls-i, 1):
                for k in balls_in_baskets(i, baskets-1):
                    yield j+k

x=[t for t in balls_in_baskets(3,3)][::-1]
for i in x:
    print(i)

我明白了：

[0, 0, 3]
[0, 1, 2]
[0, 2, 1]
[0, 3, 0]
[1, 0, 2]
[1, 1, 1]
[1, 2, 0]
[2, 0, 1]
[2, 1, 0]
[3, 0, 0]

但是，我想要这个订单：

[0, 0, 3]
[0, 1, 2]
[1, 0, 2]
[0, 2, 1]
[1, 1, 1]
[2, 0, 1]
[0, 3, 0]
[1, 2, 0]
[2, 1, 0]
[3, 0, 0]

如何实现正确的订单？

Answer 1

你已经通过列表理解消耗了你的生成器的内存效率，所以你也可以sort结果：

x = sorted(balls_in_baskets(3,3), key=lambda x: x[::-1], reverse=True)

然后打印预期的输出：

[0, 0, 3]
[0, 1, 2]
[1, 0, 2]
[0, 2, 1]
[1, 1, 1]
[2, 0, 1]
[0, 3, 0]
[1, 2, 0]
[2, 1, 0]
[3, 0, 0]

具有特定顺序的篮子中的球的计数

1 个答案: