发现总结于给定号码的所有可能子集

编程入门行业动态更新时间:2024-10-10 11:21:10

本文介绍了发现总结于给定号码的所有可能子集的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！问题描述

我正在学习Python和我有一个问题，这个看似简单的任务。

I'm learning Python and I have a problem with this seems to be simple task.

欲查找相加为一个给定数目的数字的所有可能的组合。例如：4 - > [1,1,1,1] [1,1,2] [2,2] [1,3]

I want to find all possible combination of numbers that sum up to a given number. for example: 4 -> [1,1,1,1] [1,1,2] [2,2] [1,3]

我挑选哪些生成所有可能子集（2 ^ n）和然后将溶液收率只是那些总和等于数。我的条件的问题。 code：

I pick the solution which generate all possible subsets (2^n) and then yield just those that sum is equal to the number. I have a problem with the condition. Code:

def allSum(number): #mask = [0] * number for i in xrange(2**number): subSet = [] for j in xrange(number): #if : subSet.append(j) if sum(subSet) == number: yield subSet for i in allSum(4): print i

BTW是一个不错的方法？

BTW is it a good approach?

推荐答案

下面是一些code我看到了几年前做的伎俩：

Here's some code I saw a few years ago that does the trick:

>>> def partitions(n): if n: for subpart in partitions(n-1): yield [1] + subpart if subpart and (len(subpart) < 2 or subpart[1] > subpart[0]): yield [subpart[0] + 1] + subpart[1:] else: yield [] >>> print list(partitions(4)) [[1, 1, 1, 1], [1, 1, 2], [2, 2], [1, 3], [4]]

其他参考：