leetCode-Best Time to Buy and Sell Stock with Transaction Fee
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Description:
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 = 2Output: 8Explanation: 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.
Solution:
//动态规划,到第i天有两种状态,一种是持有股票并且最大利润,有两种选择,一个是什么都不做,一个是买入股票,一种是未持有股票并且最大利润,也有两种选择,一种是什么都不做,一种是卖出股票,我们用两个数组分别保存,每次得到最优解即可,class Solution { public int maxProfit(int[] prices, int fee) { int l = prices.length; int[] hold = new int[l + 1]; //Hold the stock until day i; int[] notHold = new int[l + 1]; //Do not hold the stock until day i; hold[0] = Integer.MIN_VALUE; for (int i = 1; i <= l; i++) { hold[i] = Math.max(hold[i - 1], notHold[i - 1] - prices[i - 1] - fee); notHold[i] = Math.max(notHold[i - 1], hold[i - 1] + prices[i - 1]); } return notHold[l]; }}
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