714. Best Time to Buy and Sell Stock with Transaction Fee (Medium)

Your are given an array of integersprices, for which thei-th element is the price of a given stock on dayi; and a non-negative integerfeerepresenting 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.

https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/discuss/108870/Most-consistent-ways-of-dealing-with-the-series-of-stock-problems

Solution 1: DP O(n); O(1) 九章

1. 我们考虑最朴素的方法，对于每一天，如果当前有股票，考虑出售或者保留，如果没股票，考虑购买或者跳过，进行dfs搜索。每天都有两种操作，时间复杂度为O(2^n)
2. 如何优化呢？我们用动态规划的思想来解决这个问题，考虑每一天同时维护两种状态：拥有股票(own)状态和已经售出股票(sell)状态。用own和sell分别保留这两种状态到目前为止所拥有的最大利润。 对于sell，用前一天own状态转移，比较卖出持有股是否能得到更多的利润，即sell = max(sell , own + price - fee)， 而对于own , 我们考虑是否买新的股票更能赚钱(换言之，更优惠），own=max( own, sell-price)
3. 初始化我们要把sell设为0表示最初是sell状态且没有profit，把own设为负无穷因为最初不存在该状态，我们不希望从这个状态进行转移
4. 因为我们保存的都是最优状态，所以在买卖股票时候取max能保证最优性不变
5. 最后直接返回sell即可

    public int maxProfit(int[] prices, int fee) {
        int sell = 0; //初始化，第一天不能卖，只能买，sell表示第i天卖股票后能获得的最大利润
        int buy = -prices[0]; //buy表示第i天买股票后能获得的最大利润
        for (int i = 1; i < prices.length; i++) {
            sell = Math.max(sell, buy + prices[i] - fee); //如果卖，计算卖后能获得的最大利润
            buy = Math.max(buy, sell - prices[i]); //如果买，计算买后能获得的最大利润
        }
        return sell; //最后得到的sell一定大于buy
    }