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

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

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

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.

C++ 实现 1

推荐文章: Most consistent ways of dealing with the series of stock problems.

递推公式为:

T[i][k][0] = max(T[i-1][k][0], T[i-1][k][1] + prices[i])
T[i][k][1] = max(T[i-1][k][1], T[i-1][k][0] - prices[i] - fee)

or

T[i][k][0] = max(T[i-1][k][0], T[i-1][k][1] + prices[i] - fee)
T[i][k][1] = max(T[i-1][k][1], T[i-1][k][0] - prices[i])

代码如下:

class Solution {
public:
    int maxProfit(vector<int>& prices, int fee) {
        // s0 表示在第 i 天进行交易后, 手上有 0 支股票, 此时所获得的最大收益
        // s1 表示在第 i 天进行交易后, 手上有 1 支股票, 此时所获得的最大收益
        int s0 = 0, s1 = INT32_MIN; 
        for (auto p : prices) {
            int prev_s0 = s0;
            s0 = std::max(s0, s1 + p); // sell
            s1 = std::max(s1, prev_s0 - p - fee); // buy
        }
        return s0;
    }
};