Python中howSum函数返回结果异常求助：期望得到[2,2,2,2]却返回[[[[[],2],2],2],2]

Hey there! Let's break down why your howSum function is returning a nested list like [[[[[], 2], 2], 2], 2] instead of the flat [2, 2, 2, 2] you expect.

The Root Cause #

The problem lies in how you're combining the recursive result with the current number. Right now, your code does:

return [remainderResult, num]

This takes the existing result list (from the recursive call) and wraps it inside a new list along with num. Every recursive step adds another layer of nesting, which is why you end up with all those extra brackets.

The Fix #

Instead of wrapping the existing result, you want to append the current number to the end of the existing result list. You can do this using list concatenation:

return remainderResult + [num]

This takes the flat list from the recursive call and adds num as a new element at the end, keeping the entire structure flat.

Updated Code #

Here's the corrected version of your function:

def howSum(targetSum, numbers):
    if targetSum == 0:
        return []
    if targetSum < 0:
        return None
    for num in numbers:
        remainder = targetSum - num
        remainderResult = howSum(remainder, numbers)
        if remainderResult != None:
            return remainderResult + [num]  # Changed this line
    return None

Testing It Out #

If you run howSum(8, [2]) now, it will correctly return [2, 2, 2, 2] instead of the nested mess. Each recursive step builds onto the flat list:

• When targetSum is 0, return []
• For targetSum 2: [] + [2] → [2]
• For targetSum 4: [2] + [2] → [2, 2]
• And so on until targetSum 8 gives the full flat list.

Bonus: Optional Optimization #

If you plan to use this function with larger target sums or number lists, adding memoization (caching) will speed things up a lot by avoiding repeated calculations. Here's how you could add that:

def howSum(targetSum, numbers, memo=None):
    if memo is None:
        memo = {}
    if targetSum in memo:
        return memo[targetSum]
    if targetSum == 0:
        return []
    if targetSum < 0:
        return None
    for num in numbers:
        remainder = targetSum - num
        remainderResult = howSum(remainder, numbers, memo)
        if remainderResult != None:
            memo[targetSum] = remainderResult + [num]
            return memo[targetSum]
    memo[targetSum] = None
    return None

内容的提问来源于stack exchange，提问作者S kumar