返回子集和问题中的数组元素

Question

对于以下子集和问题的递归解决方案（see this link），如果a 中有任何子集的和等于sum 的值，则以下代码返回true

def isSubsetSum(set,n, sum) : 

    # Base Cases 
    if (sum == 0) : 
        return True
    if (n == 0 and sum != 0) : 
        return False

    # If last element is greater than sum, then ignore it 
    if (set[n - 1] > sum) : 
        return isSubsetSum(set, n - 1, sum); 

    # else, check if sum can be obtained by any of the following 
    # (a) including the last element 
    # (b) excluding the last element    
    return isSubsetSum(set, n-1, sum) or isSubsetSum(set, n-1, sum-set[n-1]) 

set = [2, 1, 14, 12, 15, 2] 
sum = 9
n = len(set) 
if (isSubsetSum(set, n, sum) == True) : 
    print("Found a subset with given sum") 
else : 
    print("No subset with given sum") 

我怎样才能返回满足sum 的a 的索引？

另外，如何使函数对数组a 中的负整数起作用。

Answer 1

这是一种可能性：

def isSubsetSum(numbers, n, x, indices):
    # Base Cases
    if (x == 0):
        return True
    if (n == 0 and x != 0):
        return False
    # If last element is greater than x, then ignore it
    if (numbers[n - 1] > x):
        return isSubsetSum(numbers, n - 1, x, indices)
    # else, check if x can be obtained by any of the following
    # (a) including the last element
    found = isSubsetSum(numbers, n - 1, x, indices)
    if found: return True
    # (b) excluding the last element
    indices.insert(0, n - 1)
    found = isSubsetSum(numbers, n - 1, x - numbers[n - 1], indices)
    if not found: indices.pop(0)
    return found

numbers = [2, 1, 4, 12, 15, 3]
x = 9
n = len(numbers)
indices = []
found = isSubsetSum(numbers, n, x, indices)
print(found)
# True
print(indices)
# [0, 2, 5]

编辑：为了更简洁的界面，您可以将前一个函数包装在另一个函数中，该函数返回成功时的索引列表，否则返回 None：

def isSubsetSum(numbers, x):
    indices = []
    found = _isSubsetSum_rec(numbers, len(numbers), x, indices)
    return indices if found else None

def _isSubsetSum_rec(numbers, n, x, indices):
    # Previous function

Answer 2

这是一种方法：如果找到，则返回子数组，而不是 True 和 False，否则返回 None：

def isSubsetSum(sset,n, ssum) : 

    # Base Cases 
    if (ssum == 0) : 
        return []
    # not found
    if (n == 0 and ssum != 0) : 
        return None

    # If last element is greater than sum, then ignore it 
    if (sset[n - 1] > ssum) : 
        return isSubsetSum(sset, n - 1, ssum); 

    # else, check if sum can be obtained by any of the following 
    # (a) including the last element 
    # (b) excluding the last element    
    a1 = isSubsetSum(sset, n-1, ssum)

    # (b) excluding last element fails
    if a1 is None:
        a2 = isSubsetSum(sset, n-1, ssum-sset[n-1])

        # (a) including last element fails
        if a2 is None:
            return None

        # (a) including last element successes
        else: return a2 + [sset[n-1]]
    else:
        return a1


sset = [2, 1, 4, 12, 15, 2] 
ssum = 9
n = len(sset) 
subset = isSubsetSum(sset, n, ssum) 
if subset is not None : 
    print(subset) 
else : 
    print("No subset with given sum") 

# [2, 1, 4, 2]

Answer 3

“简单”的方法是让索引列表成为函数的另一个参数。沿调用树向下移动时保持它。

“干净”的方法是在您返回 back 并返回成功结果的树时构建它。不是返回布尔值，而是返回索引列表； None 表示失败。

一个重要的注意事项：不要用变量名“遮蔽”一个内置类型； set 作为变量名和 具有误导性，既危险又危险。例如，试试coins。

基本情况：

if (sum == 0) : 
    return [n]
if (n == 0 and sum != 0) : 
    return None

递归：

include = isSubsetSum(coins, n-1, sum-coins[n-1]):
if include:
    return include.append(n-1)

这些是对现有代码的直接添加。我强烈建议您在线研究这个问题，看看其他人是如何解决的。将整个列表和指针作为参数传递会浪费空间和时间；相反，您可以从最后一个元素开始，然后返回，在每次调用时从列表中剪掉结尾。这将使您的索引问题变得更加简单。