K 笔交易后每股的最大利润。如何返回实际导致最大利润的交易？

Question

我有这个代码，目前返回一个大小为 n 天的列表的最大利润，有 k 笔交易。 前任。输入：价格 = [1, 5, 2, 3, 7, 6, 4, 5], k = 3,
输出 = 10

但我想将实际的 k 笔交易作为导致最大利润的买卖数组返回

for example ->[(1, 5), (2, 7),(4, 5)], instead of simply 10
#buy at 1, sell at 5...etc

我当前的代码是...

def findMaxProfit（价格，k）：

# get the number of days `n`
n = len(price)

# base case
if n <= 1:
    return 0

# profit[i][j] stores the maximum profit gained by doing
# at most `i` transactions till j'th day
profit = [[0 for x in range(n)] for y in range(k + 1)]

# fill profit[][] in a bottom-up fashion
for i in range(k + 1):
    for j in range(n):
        # profit is 0 when
        # i = 0, i.e., for 0th day
        # j = 0, i.e., no transaction is being performed

        if i == 0 or j == 0:
            profit[i][j] = 0
        else:
            max_so_far = 0
            for x in range(j):
                curr_price = price[j] - price[x] + profit[i-1][x]
                if max_so_far < curr_price:
                    max_so_far = curr_price
                    
            profit[i][j] = max(profit[i][j-1], max_so_far)
            
return profit[k][n-1]

我尝试将返回变量更改为 profit[k] ，它返回最大利润的排列 例如： 输入：价格 = [1, 5, 2, 3, 7, 6, 4, 5], k = 3

kp = findMaxProfit(price,k) -> 输出：[0, 4, 4, 5, 9, 9, 9, 10]

我可以使用 for 循环将输出的值映射到价格以获得所需的输出，但是，当数组的大小足够大时，它就不再有效了。

alpha = []
for i in range(1,len(kp)):
  if kp[i-1] != kp[i]:
    alpha.append((price[i-1],price[i]))

delta = []
delta.append(alpha[0])
for i in range(1,len(alpha)):
  if delta[-1][1] == alpha[i][0]:
    delta.append((delta.pop(-1)[0],alpha[i][1]))
  else:
    delta.append(alpha[i])

增量：[(1, 5), (2, 7), (4, 5)]

无论价目表和 k 的大小如何，有什么更好的方法来解决这个问题？

Answer 1

您使用的方法是正确的，但可以改进实现。您可以存储每笔交易的买入和卖出值，而不是存储每一天和每笔交易的利润。

这是如何实现的示例：

def findMaxProfit(price, k):
  n = len(price)

  # base case
  if n <= 1:
    return 0

  # profit[i][j] stores the maximum profit gained by doing
  # at most `i` transactions till j'th day
  profit = [[0 for x in range(n)] for y in range(k + 1)]

  # transactions[i][j] stores the buy and sell values for the
  # `i`th transaction on the `j`th day
  transactions = [[(0,0) for x in range(n)] for y in range(k + 1)]

  # fill profit[][] and transactions[][] in a bottom-up fashion
  for i in range(k + 1):
    for j in range(n):
      # profit is 0 when
      # i = 0, i.e., for 0th day
      # j = 0, i.e., no transaction is being performed

      if i == 0 or j == 0:
        profit[i][j] = 0
        transactions[i][j] = (0,0)
      else:
        max_so_far = 0
        for x in range(j):
          curr_price = price[j] - price[x] + profit[i-1][x]
          if max_so_far < curr_price:
            max_so_far = curr_price
            transactions[i][j] = (price[x], price[j])

        profit[i][j] = max(profit[i][j-1], max_so_far)

  # create the array of transactions
  transaction_array = []
  for i in range(k):
    transaction_array.append(transactions[i][n-1])

  return transaction_array

这应该返回导致最大利润的交易数组，而不管价目表和 k 的大小。

希望这可以帮助！