原题链接在这里：https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/description/
题目：
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.
题解：
T[i][k][0] 代表maximum profit that could be gained at the end of the i-th day with at most k transactions. 最后手上剩下0股stock.
递推公式就是
T[i][k][0] = max(T[i-1][k][0], T[i-1][k][1] + prices[i])
T[i][k][1] = max(T[i-1][k][1], T[i-1][k-1][0] - prices[i])
如果需要加上transaction fee就变成
T[i][k][0] = max(T[i-1][k][0], T[i-1][k][1] + prices[i])
T[i][k][1] = max(T[i-1][k][1], T[i-1][k][0] - prices[i] - fee)
买入时交fee.
最后肯定是卖掉手上的股票收益更多，所以返回T[i][k][0].
Time Complexity: O(n). n = prices.length.
Space: O(1).
AC Java:
1 class Solution {
2 public int maxProfit(int[] prices, int fee) {
3 int tIk0 = 0;
4 int tIk1 = Integer.MIN_VALUE;
5 for(int price : prices){
6 int preTransactionIk0 = tIk0;
7 tIk0 = Math.max(tIk0, tIk1+price);
8 tIk1 = Math.max(tIk1, tIk0-price-fee);
9 }
10 return tIk0;
11 }
12 }
类似Best Time to Buy and Sell Stock II.
买卖股票类题目都可以套用这个思路.
Reference: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/discuss/108870/
原文地址：https://www.cnblogs.com/Dylan-Java-NYC/p/8131710.html