Leetcode Problem 417

2018/12/9

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

问题描述

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.

测试样例

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.

问题分析

本题难度为Medium！属于动态规划问题，已给出的函数定义为

class Solution:
    def maxProfit(self, prices, fee):
        """
        :type prices: List[int]
        :type fee: int
        :rtype: int
        """

本题可以采用动态规划的解决办法；
dp[i][0]：第i天手中没有股票时能赚的最高价；
dp[i][1]：第i天手中有股票时所赚的最高价；

当第i天手里没有股票时，这一天有两个情况，第一种是在今天卖出了股票，从而一共赚取了dp[i-1][1]+prices[i]-fee，第二种情况是昨天就没有股票，今天只是仍然没有股票而已，这个时候dp[i][0]=dp[i-1][0];同理可以推导出dp[i][1]的转移方程;
最终答案就是dp[n][0]，因为必然要让手里的钱更多，肯定是手里没有股票的时候。
转移方程为：
dp[i][0] = max(dp[i-1][1]+prices[i]-fee, dp[i-1][0]);
dp[i][1] = max(dp[i-1][0]-prices[i], dp[i-1][1]);

代码实现

python3

# coding: utf-8
class Solution:
    def maxProfit(self, prices, fee):
        """
        :type prices: List[int]
        :type fee: int
        :rtype: int
        """
        if len(prices) == 0: return 0
        dp = [[0,0] for i in range(len(prices))]
        dp[0][0] = 0
        dp[0][1] = -prices[0]
        for i in range(1, len(prices)):
            dp[i][0] = max(dp[i-1][0], dp[i-1][1]+prices[i]-fee)
            dp[i][1] = max(dp[i-1][1], dp[i-1][0]-prices[i])
        return dp[-1][0]