Question

我想从Finding all possible combinations of numbers to reach a given sum修改subset_sum() python函数，以便：

它允许重复（排列）而不是组合
它仅考虑给定长度的排列

我已经成功完成了＃2，但是我需要＃1的帮助：

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

    # check if the partial sum is equals to target
    if s == target and len(partial) == length:
        print(f"sum({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, length, partial + [n])

所需的输出应为：

>>> subset_sum([3,9,8,4,5,7,10],target=15,length=3)
sum([3, 8, 4])=15
sum([3, 4, 8])=15
sum([4, 3, 8])=15
sum([4, 8, 3])=15
sum([8, 3, 4])=15
sum([8, 4, 3])=15
sum([3, 5, 7])=15
sum([3, 7, 5])=15
sum([5, 3, 7])=15
sum([5, 7, 3])=15
sum([7, 3, 5])=15
sum([7, 5, 3])=15

Answer 1

由于您已经解决了在每个等效组中标识一个解决方案的问题，因此我的建议是：不要不更改该算法。而是利用itertools.permutations生成这些项目：

return list(itertools.permutations(numbers))

Answer 2

做到这一点：

import itertools

numbers = [3,9,8,4,5,7,10]
length = 3
target = 15

iterable = itertools.permutations(numbers,length)
predicate = lambda x: (sum(x) == target)
vals = filter(predicate,iterable)
list(vals)

或单线：

vals = [x for x in itertools.permutations(numbers,length) if sum(x) == target]

结果：

[(3, 8, 4),
 (3, 4, 8),
 (3, 5, 7),
 (3, 7, 5),
 (8, 3, 4),
 (8, 4, 3),
 (4, 3, 8),
 (4, 8, 3),
 (5, 3, 7),
 (5, 7, 3),
 (7, 3, 5),
 (7, 5, 3)]

查找固定长度的数字的所有可能排列以达到给定的总和

2 个答案: