- Best Time to Buy and Sell Stock with Transaction Fee
Medium
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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.
思路
有两个变量：
- crash记录，手上没有股票的最大收益
- hold记录，手上持有股票的最大收益
每次遇到新的股票，都会更新crash和hold
- cash = Math.max(cash, hold + prices[i] - fee)，以前手上没有股票的收益高，还是当前卖掉股票的收益高
- hold = Math.max(hold, cash - prices[i])，以前手上持有股票的收益高，还是现在买进股票的收益高
java
class Solution {
public int maxProfit(int[] prices, int fee) {
int cash = 0, hold = -prices[0];
for (int i = 1; i < prices.length; i++) {
cash = Math.max(cash, hold + prices[i] - fee);
hold = Math.max(hold, cash - prices[i]);
}
return cash;
}
}
python
class Solution:
def maxProfit(self, prices: List[int], fee: int) -> int:
crash, hold = 0, -prices[0]
for i in range(1, len(prices)):
crash = max(crash, hold + prices[i] - fee)
hold = max(hold, crash - prices[i])
return crash
"""
:type prices: List[int]
:type fee: int
:rtype: int
"""