【leetcode】714. Best Time to Buy and Sell Stock with Transaction Fee

题目如下：

Your are given an array of integers prices, for which the i-th element is the price of a given stock on day i; and a non-negative integer feerepresenting a transaction fee.

You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction. You may not buy more than 1 share of a stock at a time (ie. you must sell the stock share before you buy again.)

Return the maximum profit you can make.

Example 1:

Input: prices = [1, 3, 2, 8, 4, 9], fee = 2
Output: 8
Explanation: 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] = 9

The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.

 

Note:

• 0 < prices.length <= 50000.
• 0 < prices[i] < 50000.
• 0 <= fee < 50000.

解题思路：对于任意一天i来说，可以有跳过(cooldown包含在内)/买入/卖出三种操作，记dp[i][0]，dp[i][1]，dp[i][2]为第i天是做这三种操作时候可以获得的最大收益。很显然 dp[i][0] = max(dp[i-1][0],dp[i-1][1],dp[i-1][2])；dp[i][1]的情况分为第一次买入/非第一次买入，所以有 dp[i][1] = max(dp[i][1] ，-price[i]，dp[j][2] - prices[i] ) (j<i) ，-prices[i]表示第一次买入，在当前节点获得的收益是负数，因为钱已经变成了股票，dp[j][2] - prices[i]表示本次买入是在第j天卖出后的后序操作，中间不存在其他的交易，这里还需要减去fee，因为每次卖出的时候需要收取手续费；同理dp[i][2] = max(dp[i][2]，dp[j][1] + prices[i]-fee)，这是因为卖出是要在买入之后。当然，对于每个i来说，并不需要去比较0~j天的所有数据，只要记录之前出现过的dp[j][1]和dp[j][2]的最大值即可。最后的结果是 max(0, dp[-1][0], dp[-1][2])，因为最后一天要么跳过，要么卖出，不会再有买入的操作。

代码如下：

class Solution(object):
    def maxProfit(self, prices, fee):
        """
        :type prices: List[int]
        :type fee: int
        :rtype: int
        """
        if len(prices) <= 1:
            return 0
        dp = []
        for i in prices:
            dp.append([-float('inf'), -float('inf'), -float('inf')])  # 0:do nothing, 1:buy ,2:sell
        dp[0][1] = -prices[0]
        #dp[1][1] = -prices[1]
        #dp[1][2] = prices[1] - prices[0] - fee

        max_buy = max(dp[0][1], dp[1][1])
        max_sell = -float('inf')
        for i in range(1, len(dp)):
            dp[i][0] = max(dp[i - 1][0], dp[i - 1][1], dp[i - 1][2])
            dp[i][2] = max(dp[i][2], max_buy + prices[i]-fee)
            dp[i][1] = max(dp[i][1], -prices[i], max_sell - prices[i])
            max_sell = max(max_sell, dp[i][2])  
            max_buy =  max(max_buy, dp[i][1])
        #print dp
        return max(0, dp[-1][0], dp[-1][2])