0714. Best Time to Buy and Sell Stock with Transaction Fee (M)

Best Time to Buy and Sell Stock with Transaction Fee (M)

题目

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 * 10^4
• 0 < prices[i], fee < 5 * 10^4

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题意

股票买卖问题。给定每一天的股票价格以及相应规则：同一天只能买或卖，卖股票必须在买股票之后，可以执行多次买卖交易，但每次卖股票后要扣除一定的手续费。

思路

和 0309.Best Time to Buy and Sell Stock with Cooldown (M) 解法基本一致。动态规划解决。

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代码实现

Java

class Solution {
    public int maxProfit(int[] prices, int fee) {
        int[] hold = new int[prices.length];
        int[] sold = new int[prices.length];

        hold[0] = -prices[0];
        sold[0] = 0;

        for (int i = 1; i < prices.length; i++) {
            hold[i] = Math.max(hold[i - 1], sold[i - 1] - prices[i]);
            sold[i] = Math.max(sold[i - 1], hold[i - 1] + prices[i] - fee);
        }

        return sold[prices.length - 1];
    }
}