LeetCode | 0714. Best Time to Bu

LeetCode 0714. Best Time to Buy and Sell Stock with Transaction Fee买卖股票的最佳时机含手续费【Medium】【Python】【动态规划】

Problem

LeetCode

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] = 1Selling at prices[3] = 8Buying at prices[4] = 4Selling at prices[5] = 9The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.

Note:

0 < prices.length <= 50000.

0 < prices[i] < 50000.

0 <= fee < 50000.

问题

力扣

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.

思路

动态规划

相当于在 LeetCode 0122 基础上加了手续费。

找到状态方程

dp[i][k][0] = max(dp[i-1][k][0], dp[i-1][k][1] + prices[i] - fee)
解释：昨天没有股票，昨天有股票今天卖出，同时减去交易费用（交易费用记在买或卖都可以）

dp[i][k][1] = max(dp[i-1][k][1], dp[i-1][k][0] - prices[i])
解释：昨天有股票，昨天没有股票今天买入

base case：
dp[-1][k][0] = dp[i][k][0] = 0
dp[-1][k][1] = dp[i][k][1] = -inf

k = +inf
因为 k 为正无穷，那么可以把 k 和 k-1 看成是一样的。
buy+sell = 一次完整的交易，这里把 sell 看成一次交易，所以第一行是 k-1。
dp[i][k][0] = max(dp[i-1][k][0], dp[i-1][k-1][1] + prices[i] - fee)
            = max(dp[i-1][k][0], dp[i-1][k][1] + prices[i] - fee)
dp[i][k][1] = max(dp[i-1][k][1], dp[i-1][k][0] - prices[i])

所以 k 对状态转移没有影响：
dp[i][0] = max(dp[i-1][0], dp[i-1][1] + prices[i] - fee)
dp[i][1] = max(dp[i-1][1], dp[i-1][0] - prices[i])

i = 0 时，dp[i-1] 不合法。
dp[0][0] = max(dp[-1][0], dp[-1][1] + prices[i] - fee)
         = max(0, -infinity + prices[i] - fee)
         = 0
dp[0][1] = max(dp[-1][1], dp[-1][0] - prices[i])
         = max(-infinity, 0 - prices[i]) 
         = -prices[i]

空间复杂度: O(1)

Python3代码

class Solution:
    def maxProfit(self, prices: List[int], fee: int) -> int:
        dp_i_0 = 0
        dp_i_1 = float('-inf')  # 负无穷
        for i in range(len(prices)):
            temp = dp_i_0
            # 昨天没有股票，昨天有股票今天卖出，同时减去交易费用
            dp_i_0 = max(dp_i_0, dp_i_1 + prices[i] - fee)  # dp_i_0 和 dp_i_1 可以看成是变量，存储的都是上一次即昨天的值
            # 昨天有股票，昨天没有股票今天买入
            dp_i_1 = max(dp_i_1, temp - prices[i])
        return dp_i_0   

代码地址

GitHub链接

参考

一个方法团灭 6 道股票问题