查找固定長度數字的所有可能排列以達到給定的總和

Question

我想修改subset_sum() python 函數從查找所有可能的數字組合以達到給定的總和，以便：

1. 它允許重復（排列）而不是組合
2. 它只考慮給定長度的排列

我已經成功完成了#2，但我需要#1 的幫助：

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]) 

所需的輸出應該是：

>>> 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

既然你已經解決了識別每個等價組中的一個解決方案的問題，我的建議是：不要改變該算法。 相反，利用itertools.permutations來生成這些項目：

return list(itertools.permutations(numbers))

Answer 2

這樣做：

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)

或單線：

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

結果：

[(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)]