{leetcode}/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/[LeetCode - Best Time to Buy and Sell Stock with Transaction Fee^]
Your are given an array of integers prices, for which the ith 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.
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.
针对 一个方法团灭 6 道股票问题 - 最佳买卖股票时机含冷冻期 - 力扣（LeetCode） 这个解题框架，进行小试牛刀。
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</li>. Selling at prices[3] = 8</li>. Buying at prices[4] = 4</li>. Selling at prices[5] = 9</li>The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.
Note:
. 0 < prices.length ⇐ 50000.</li>
. 0 < prices[i] < 50000.</li>
. 0 ⇐ fee < 50000.</li>
link:{sourcedir}/_0714_BestTimeToBuyAndSellStockWithTransactionFee.java[role=include]