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

 

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

 

题目

 

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 fee representing 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] = 1Selling at prices[3] = 8Buying at prices[4] = 4Selling at prices[5] = 9The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.

Note:0 < prices.length <= 50000.0 < prices[i] < 50000.0 <= fee < 50000.

分析

给定一个价格数组，第i个元素表示第i天一个给定股票的价格；fee是非负的交易费用，每次交易卖方必须要支付。

本题采用动态规划（Dynamic Programming）的方法，时间复杂度O(N)，空间复杂度O(1).

定义两个变量sold和hold：

sold[i] 表示第i天卖掉手中股票所能取得的最大收益，即该天结束后手中不持有股票（保持i-1天不持有的状态/卖掉手中的股票）

hold[i] 表示第i天拥有某个股票时所能获得的最大收益，即该天结束后手中持有股票（保持i-1天持有的状态/买入新股票）

卖出股票时出交易费：

sold[i]=max(sold[i-1], hold[i-1]+p[i]-fee)    #表示取已获得收益与当前卖掉股票的最大值
hold[i]=max(hold[i-1], sold[i-1]-p[i])    #表示取继续持有或买入股票可能获得收益的最大值

买入股票时出交易费：

sold=max(sold,hold+price[i])
hold=max(hold,sold-price[i]-fee)

最后返回遍历完最后一个prices元素之后，手中不持有股票的收益。

完整代码

class Solution(object):
    def maxProfit(self, prices, fee):
        """
        :type prices: List[int]
        :type fee: int
        :rtype: int
        """
        sold,hold=0,-prices[0]  #第0天买入股票，sold为当前收益
        for i in xrange(len(prices)):
            sold = max(sold, hold+prices[i]-fee)    #卖掉股票
            hold = max(hold, sold-prices[i])    #买入股票
        return sold  #输出sold，因为最后一天手中不持有股票

运行速度快于DP max()的方法，采用变量state记录用当前价格卖掉/买入股票所能获得的收益:

    如果state>fee(大于额外的交易费用fee)即可进行交易；如果state<0，当前价格下不进行交易。

class Solution(object):
    def maxProfit(self, prices, fee):
        """
        :type prices: List[int]
        :type fee: int
        :rtype: int
        """
        state=profit=0  #初始化总收益profit，当前价格下进行交易的损益state
        last_price=prices[0]   #前一天的价格记为last_price
        for price in prices[1:]:    #从第二天的价格开始遍历
            state+=price-last_price #若按当前价格卖掉，相比于前一天的损益
            if state>fee:   #若纯收益为正，则卖掉，更新收益并state归零
                profit+=state-fee
                state=0
            elif state<0:   #若亏本，state归零
                state=0
            last_price=price#记录当前价格作为下一次循环的前一天价格
        return profit