【LeetCode】714. Best Time to Buy and Sell Stock with Transaction Fee 解题报告（Python & C++）

时间：2021-02-07 23:54:13

作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

题目地址：https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/description/

题目描述

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] = 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.

题目大意

给出一系列的股票交易价格，每次股票交易会有交易fee，求买卖股票能获得的最大的收益。

解题方法

动态规划

可以使用dp的方法去做。该dp使用了两个数字，cash和hold。解释如下：

  1. cash 该天结束手里没有股票的情况下，已经获得的最大收益
  2. hold 该天结束手里有股票的情况下，已经获得的最大收益

所以转移状态分析如下：

cash 更新的策略是：既然今天结束之后手里没有股票，那么可能是今天没买（保持昨天的状态），也可能是今天把股票卖出了

hold 更新的策略是：今天今天结束之后手里有股票，那么可能是今天没卖（保持昨天的状态），也可能是今天买了股票

class Solution:

    def maxProfit(self, prices, fee):

        """

        :type prices: List[int]

        :type fee: int

        :rtype: int

        """

        cash = 0

        hold = -prices[0]

        for i in range(1, len(prices)):

            cash = max(cash, hold + prices[i] - fee)

            hold = max(hold, cash - prices[i])

        return cash

使用C++代码如下：

class Solution {

public:

    int maxProfit(vector<int>& prices, int fee) {

        const int N = prices.size();

        vector<int> cash(N, 0);

        vector<int> hold(N, 0);

        cash[0] = 0;

        hold[0] = -prices[0];

        for (int i = 1; i < N; i ++) {

            cash[i] = max(cash[i - 1], prices[i] + hold[i - 1] - fee);

            hold[i] = max(hold[i - 1], cash[i - 1] - prices[i]);

        }

        return cash[N - 1];

    }

};

优化空间复杂度到O(1)的代码如下：

class Solution {

public:

    int maxProfit(vector<int>& prices, int fee) {

        const int N = prices.size();

        // max profit if today don't have stock

        int cash = 0;

        // max profit if today have stock

        int hold = -prices[0];

        for (int i = 1; i < N; ++i) {

            cash = max(cash, prices[i] + hold - fee);

            hold = max(hold, cash - prices[i]);

        }

        return cash;

    }

};

日期

2018 年 4 月 10 日 —— 风很大，很舒服～～
2018 年 12 月 5 日 —— 周三啦！
2019 年 2 月 22 日 —— 这周结束了

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