使用递归查找总和相等的情况

Question

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]) 

The above code is supposed to check how many times a certain target sum can be obtained from a list.

How may I modify it to use recursion to check the number of times the sums of any combination of numbers in the list are equal? For example, [1, 2, 3] should give an output of 2 since 1+2=3 and 3= 2+1 and [1, 1, 2, 3, 5] should give 4.

Answer 1

Something like this should work:

def subset_sum(target, remaining, partial=()):
    remaining = tuple(remaining)
    if remaining == ():
        return 1 if sum(partial) == target else 0
    else:
        return (subset_sum(target, partial, remaining[1:]) + 
                subset_sum(target, partial + remaining[:1], remaining[1:]))

>>> subset_sum(5, [1,1,2,3,5])
3
>>> subset_sum(2, [1,1,2,3,5])
2

We use tuples here do we don't have to worry about multiple references to the same list interfering with each other. Tuples are immutable so any "modifications" won't spill into other instances.