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
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
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 
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