算法: 买卖股票并且有手续费714. Best Time to Buy and Sell Stock with Transaction Fee

Posted 2021-12-24 AI架构师易筋

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

You are given an array prices where prices[i] is the price of a given stock on the ith day, and an integer fee representing a transaction fee.

Find the maximum profit you can achieve. You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction.

Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

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.

Example 2:

Input: prices = [1,3,7,5,10,3], fee = 3
Output: 6

Constraints:

• 1 <= prices.length <= 5 * 104
• 1 <= prices[i] < 5 * 104
• 0 <= fee < 5 * 104

动态规划解法：

给定 any day I，其最大利润状态归结为以下两种状态之一：

（1）购买状态：

buy[i]代表了最大的利润，在day i在买状态，因为你采取的最后一个动作是一个买入的动作在day K其中K<=i。并且您有权在 出售day i+1，或者什么也不做。

（2）销售情况：

sell[i]代表最大的利润在day i在卖出状态，因为你采取的最后一个动作是一个卖出动作的day K，在那里K<=i。并且您有权在 购买day i+1，或者什么都不做。

让我们从基本案例开始。

基本情况：

我们可以从购买状态开始，这意味着我们在 购买股票day 0。

buy[0]=-prices[0];

或者我们可以从卖出状态开始，这意味着我们在day 0.
鉴于我们在第 0 天手头没有任何库存，我们将卖出状态设置为 0

sell[0]=0。;

状态转换：

在day i，我们可以买入股票（从之前的卖出状态）或什么都不做（从之前的买入状态）： 或者 在，我们可以卖出股票（从之前的买入状态）或保持持有（从之前的卖出状态）：

buy[i] = Math.max(buy[i - 1], sell[i - 1] - prices[i]);

day i

sell[i] = Math.max(sell[i - 1], buy[i - 1] + prices[i]);

最后：

我们将返回sell[last_day]作为我们的结果，它代表最后一天的最大利润，因为您在最后一天之前的任何一天采取了卖出行动。

我们可以在买入或卖出状态收取交易费。

所以这里来了我们的两个解决方案：

方案1——卖出股票时支付费用：

  public int maxProfit(int[] prices, int fee) 
        if (prices.length <= 1) return 0;
        int days = prices.length, buy[] = new int[days], sell[] = new int[days];
        buy[0]=-prices[0];
        for (int i = 1; i<days; i++) 
            buy[i] = Math.max(buy[i - 1], sell[i - 1] - prices[i]); // keep the same as day i-1, or buy from sell status at day i-1
            sell[i] = Math.max(sell[i - 1], buy[i - 1] + prices[i] - fee); // keep the same as day i-1, or sell from buy status at day i-1
        
        return sell[days - 1];
    

方案2——购买股票时支付费用：

public int maxProfit(int[] prices, int fee) 
        if (prices.length <= 1) return 0;
        int days = prices.length, buy[] = new int[days], sell[] = new int[days];
        buy[0]=-prices[0]-fee;
        for (int i = 1; i<days; i++) 
            buy[i] = Math.max(buy[i - 1], sell[i - 1] - prices[i] - fee); // keep the same as day i-1, or buy from sell status at day i-1
            sell[i] = Math.max(sell[i - 1], buy[i - 1] + prices[i]); // keep the same as day i-1, or sell from buy status at day i-1
        
        return sell[days - 1];
    

参考

https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/discuss/108871/2-solutions-2-states-DP-solutions-clear-explanation!