LeetCode714. Best Time to Buy and Sell Stock with Transaction Fee

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本文介绍了一种在考虑交易手续费的情况下实现股票买卖最大利润的算法。通过动态调整买入和卖出状态，确保每笔交易都能获得正收益。示例中使用了价格波动和固定手续费的数据，展示了如何计算最大可能利润。

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原题：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：
sell = max(sell, buy+prices[i])
buy = max(buy, sell-prices[i]-fee)
令sell为还未购买时手中的金额，buy为还未售出时的手中的金额
每次sell都和在当前售出时所能获得的金额相比较，如果buy+prices[i]更大，说明现在售出是有利可图的，更新sell。
每次buy都和当前金额减去当前价格以及购买费用比较，更新为更大的一方，因为buy越大，后面sell获利越大。
一般而言当sell更新时，buy不会立即更新，因为此时sell-prices[i]-fee = buy-fee < buy，这样就不会产生购入和售出的冲突。

class Solution {
public:
    int maxProfit(vector<int>& prices, int fee) {
        int sell = 0, buy = -500000;
        for (int i = 0; i < prices.size(); i++) {
            sell = max(sell, buy+prices[i]);
            buy = max(buy, sell-prices[i]-fee);
        }
        return sell;
    }
};