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

题目 #

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 #

给定一个整数数组 prices，其中第 i 个元素代表了第 i 天的股票价格 ；非负整数 fee 代表了交易股票的手续费用。你可以无限次地完成交易，但是你每次交易都需要付手续费。如果你已经购买了一个股票，在卖出它之前你就不能再继续购买股票了。要求返回获得利润的最大值。

解题思路 #

• 给定一个数组，表示一支股票在每一天的价格。设计一个交易算法，在这些天进行自动交易，要求：每一天只能进行一次操作；在买完股票后，必须卖了股票，才能再次买入；每次卖了股票以后，需要缴纳一部分的手续费。问如何交易，能让利润最大？
• 这一题是第 121 题、第 122 题、第 309 题的变种题。
• 这一题的解题思路是 DP，需要维护买和卖的两种状态。buy[i] 代表第 i 天买入的最大收益，sell[i] 代表第 i 天卖出的最大收益，状态转移方程是 buy[i] = max(buy[i-1], sell[i-1]-prices[i])，sell[i] = max(sell[i-1], buy[i-1]+prices[i]-fee)。

代码 #

package leetcode

import (
	"math"
)

// 解法一 模拟 DP
func maxProfit714(prices []int, fee int) int {
	if len(prices) <= 1 {
		return 0
	}
	buy, sell := make([]int, len(prices)), make([]int, len(prices))
	for i := range buy {
		buy[i] = math.MinInt64
	}
	buy[0] = -prices[0]
	for i := 1; i < len(prices); i++ {
		buy[i] = max(buy[i-1], sell[i-1]-prices[i])
		sell[i] = max(sell[i-1], buy[i-1]+prices[i]-fee)
	}
	return sell[len(sell)-1]
}

// 解法二 优化辅助空间的 DP
func maxProfit714_1(prices []int, fee int) int {
	sell, buy := 0, -prices[0]
	for i := 1; i < len(prices); i++ {
		sell = max(sell, buy+prices[i]-fee)
		buy = max(buy, sell-prices[i])
	}
	return sell
}