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【Python】
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
