[Swift]LeetCode714. 买卖股票的最佳时机含手续费 | Best Time to Buy and Sell Stock with Transaction Fee

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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.

---

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。

返回获得利润的最大值。

示例 1:

输入: prices = [1, 3, 2, 8, 4, 9], fee = 2
输出: 8
解释: 能够达到的最大利润:  
在此处买入 prices[0] = 1
在此处卖出 prices[3] = 8
在此处买入 prices[4] = 4
在此处卖出 prices[5] = 9
总利润: ((8 - 1) - 2) + ((9 - 4) - 2) = 8.

注意:

• 0 < prices.length <= 50000.
• 0 < prices[i] < 50000.
• 0 <= fee < 50000.

---

Runtime: 796 ms

Memory Usage: 19.8 MB

class Solution {
    func maxProfit(_ prices: [Int], _ fee: Int) -> Int {
        var sold:Int = 0
        var hold:Int = -prices[0]
        for price in prices
        {
            var t:Int = sold
            sold = max(sold, hold + price - fee)
            hold = max(hold, t - price)
        }
        return sold
    }
}

---

880ms

class Solution {
    func maxProfit(_ prices: [Int], _ fee: Int) -> Int {        
        var cash = 0, hold = -prices[0]
        for i in prices.indices {
            cash = max(cash, hold + prices[i] - fee)
            hold = max(hold, cash - prices[i])
        }
        return cash
    }

    func dpMaxProfit(_ prices: [Int], _ fee: Int, _ index: Int, _ end: Int, _ profit: Int, _ buyed: Bool) -> Int {

        if index >= end {
            return profit
        }

        if buyed {
            let sum2 = dpMaxProfit(prices, fee, index+1, end, profit, false)
            let sum3 = dpMaxProfit(prices, fee, index+1, end, profit - fee + prices[index], false)
            return max(sum2, sum3)
        }
        else {
            let sum1 = dpMaxProfit(prices, fee, index+1, end, profit - prices[index], true)
            let sum2 = dpMaxProfit(prices, fee, index+1, end, profit, false)
            return max(sum1, sum2)
        }
    }
}

---

948ms

class Solution {
    func maxProfit(_ prices: [Int], _ fee: Int) -> Int {        
        var totalProfit = 0, currProfit = 0, buy = 0, sell = 0
        for i in 0..<prices.count {
            if prices[i] < prices[buy] || prices[sell] - prices[i] > fee {
                buy = i
                sell = i
                totalProfit += currProfit
                currProfit = 0
            } else if prices[i] > prices[sell] && prices[i] - prices[buy] > fee {
                sell = i
                currProfit = prices[sell] - prices[buy] - fee
            }
        }
        totalProfit += currProfit
        return totalProfit
    }
}