Python：递归函数。如何返回targetsum的所有子集

Question

我的代码没有显示最短的子集，例如[7]，或者它没有读取所有的子集[7]，[3,4]，以返回最短的子集。可以解释为什么只有一组结果是返回的，我应该如何修改它以显示所有子集？谢谢

下面是我想要遵循的代码的图片

def howsum(targetsum,numbers,combo=None):
    
    if combo == None:
        combo = list()
    if targetsum == 0: return [ ]
    if targetsum < 0: return None
    shortcombo = None
    
    for number in numbers:
        remainder = targetsum - number      
        
        combo = howsum(remainder,numbers,combo)
        if combo != None:
            combo.append(number)
            if shortcombo == None or len(shortcombo) > len(combo):
                shortcombo = combo
                return shortcombo
                                    
    return shortcombo
     
print(howsum(7,[4,3,7]))

Answer 1

我不确定这一点，但在python中定义函数并给出两组代码时，调用函数时实际上并没有执行第二组代码。我在其他项目中遇到了同样的问题。我没有得到答案，只是说我在代码中犯了一些错误

Answer 2

增加了记忆，但是里面有1组num。结果乱七八糟

def best_sum(target_sum, numbers,memo=None):
    if memo == None:
      memo = dict()
    
    if target_sum in memo:
      return memo[target_sum]
  
    if target_sum == 0: return []
    if target_sum < 0: return None
    
    shortest_combination = None

    for num in numbers:
        remainder = target_sum - num
        remainder_combination = best_sum(remainder, numbers,memo)
        if remainder_combination != None:
            remainder_combination.append(num)
            combination = remainder_combination
            if shortest_combination == None or len(combination) < len(shortest_combination):
                shortest_combination = combination
    memo[target_sum] = shortest_combination
    return shortest_combination

print(best_sum(100,[5,1,25]))

此代码由DarylG works发布。通过将.append更改为[*list,var]，我不明白为什么函数append和*

def best_sum(target_sum, numbers,memo=None):
    if memo == None:
      memo = dict()
    
    if target_sum in memo:
      return memo[target_sum]
  
    if target_sum == 0: return []
    if target_sum < 0: return None
    
    shortest_combination = None

    for num in numbers:
        remainder = target_sum - num
        remainder_combination = best_sum(remainder, numbers,memo)
        if remainder_combination != None:
            combination = [*remainder_combination,num] #this line from append to *
            if shortest_combination == None or len(combination) < len(shortest_combination):
                shortest_combination = combination
    memo[target_sum] = shortest_combination
    return shortest_combination

print(best_sum(100,[5,1,25]))

Answer 3

编写的代码与原始JavaScript非常匹配

虽然JavaScript名称可以工作，但我对函数和变量名称进行了重构，使之与Python style一致，即：

• 函数名应该是小写的，并根据需要用下划线分隔单词，以提高可读性
• 变量名遵循与函数名相同的约定

代码

def best_sum(target_sum, numbers):
    if target_sum == 0: return []
    if target_sum < 0: return None
    
    shortest_combination = None

    for num in numbers:
        remainder = target_sum - num
        remainder_combination = best_sum(remainder, numbers)
        if remainder_combination != None:
            combination = [*remainder_combination, num] # Python * equivalent to JavaSscript ...
            if shortest_combination == None or len(combination) < len(shortest_combination):
                shortest_combination = combination
    
    return shortest_combination

测试

print(bestSum(7, [3, 4, 7])) # Output: [7]

使用记忆（即缓存）

def best_sum(target_sum, numbers, memo = None):
    if memo is None:
        memo = {0:[]}
    if target_sum < 0:
        return None
    if target_sum in memo:
        return memo[target_sum]

    shortest_combination = None

    for num in numbers:
        remainder = target_sum - num

        remainder_combination = best_sum(remainder, numbers, memo)
        if remainder_combination != None:
            combination = [*remainder_combination, num] # Python * equivalent to JavaSscript ...
            if shortest_combination == None or len(combination) < len(shortest_combination):
                shortest_combination = combination

    memo[target_sum] = shortest_combination

    return memo[target_sum]

print(best_sum(7, [3, 4, 7]))  # Output: 7
# Following very slow on non-memoized version
print(best_sum(100,[10,1,25])) # Output: [25, 25, 25, 25]