LeetCode 0714. Best Time to Buy and Sell Stock with Transaction Fee買賣股票的最佳時(shí)機(jī)含手續(xù)費(fèi)【Medium】【Python】【動(dòng)態(tài)規(guī)劃】
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.
問題
給定一個(gè)整數(shù)數(shù)組 prices干签，其中第 i 個(gè)元素代表了第 i 天的股票價(jià)格 殴穴；非負(fù)整數(shù) fee 代表了交易股票的手續(xù)費(fèi)用突诬。
你可以無限次地完成交易，但是你每次交易都需要付手續(xù)費(fèi)。如果你已經(jīng)購買了一個(gè)股票僚纷，在賣出它之前你就不能再繼續(xù)購買股票了。
返回獲得利潤的最大值。
示例 1:
輸入: prices = [1, 3, 2, 8, 4, 9], fee = 2
輸出: 8
解釋: 能夠達(dá)到的最大利潤:
在此處買入 prices[0] = 1
在此處賣出 prices[3] = 8
在此處買入 prices[4] = 4
在此處賣出 prices[5] = 9
總利潤: ((8 - 1) - 2) + ((9 - 4) - 2) = 8.
注意:
-
0 < prices.length <= 50000. -
0 < prices[i] < 50000. -
0 <= fee < 50000.
思路
動(dòng)態(tài)規(guī)劃
相當(dāng)于在 LeetCode 0122 基礎(chǔ)上加了手續(xù)費(fèi)疚察。
找到狀態(tài)方程
dp[i][k][0] = max(dp[i-1][k][0], dp[i-1][k][1] + prices[i] - fee)
解釋：昨天沒有股票，昨天有股票今天賣出仇奶，同時(shí)減去交易費(fèi)用（交易費(fèi)用記在買或賣都可以）
dp[i][k][1] = max(dp[i-1][k][1], dp[i-1][k][0] - prices[i])
解釋：昨天有股票貌嫡，昨天沒有股票今天買入
base case：
dp[-1][k][0] = dp[i][k][0] = 0
dp[-1][k][1] = dp[i][k][1] = -inf
k = +inf
因?yàn)?k 為正無窮，那么可以把 k 和 k-1 看成是一樣的该溯。
buy+sell = 一次完整的交易岛抄，這里把 sell 看成一次交易，所以第一行是 k-1狈茉。
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 對(duì)狀態(tài)轉(zhuǎn)移沒有影響：
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 時(shí)夫椭，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]
空間復(fù)雜度: O(1)
Python3代碼
class Solution:
def maxProfit(self, prices: List[int], fee: int) -> int:
dp_i_0 = 0
dp_i_1 = float('-inf') # 負(fù)無窮
for i in range(len(prices)):
temp = dp_i_0
# 昨天沒有股票氯庆，昨天有股票今天賣出蹭秋，同時(shí)減去交易費(fèi)用
dp_i_0 = max(dp_i_0, dp_i_1 + prices[i] - fee) # dp_i_0 和 dp_i_1 可以看成是變量扰付，存儲(chǔ)的都是上一次即昨天的值
# 昨天有股票，昨天沒有股票今天買入
dp_i_1 = max(dp_i_1, temp - prices[i])
return dp_i_0
