動機

與309、188一起看效果更佳

Problem

You are given an array prices where prices[i] is the price of a given stock on the ith day, and an integer fee representing a transaction fee.

Find the maximum profit you can achieve. You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction.

Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

 

Example 1:

Input: prices = [1,3,2,8,4,9], fee = 2Output: 8Explanation: The maximum profit can be achieved by:- Buying at prices[0] = 1- Selling at prices[3] = 8- Buying at prices[4] = 4- Selling at prices[5] = 9The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.

Example 2:

Input: prices = [1,3,7,5,10,3], fee = 3Output: 6

 

Constraints:

  • 1 <= prices.length <= 5 * 104
  • 1 <= prices[i] < 5 * 104
  • 0 <= fee < 5 * 104

Sol

現在是在賣的時候要手續費

所以原本的式子，在賣的時候要扣$$

buy[i] = sell[i-1] - prices[i] | buy[i-1]
sell[i] = buy[i-1] + prices[i] | sell[i-1]

buy[i] = sell[i-1] - prices[i] | buy[i-1]
sell[i] = buy[i-1] + prices[i] - fee | sell[i-1]

class Solution:
    @functools.cache
    def buy(self,i):
        if i < 0:
            return 0
        elif i == 0:
            return -self.prices[i]
        else:
            return max(self.sell(i-1)-self.prices[i], self.buy(i-1))
    @functools.cache
    def sell(self,i):
        if i <= 0:
            return 0
        else:
            return max(self.buy(i-1)+self.prices[i] - self.fee,self.sell(i-1))
    def maxProfit(self, prices: List[int], fee: int) -> int:
        self.fee = fee
        self.prices = prices
        return self.sell(len(prices)-1)