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

原题：https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/description/

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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 = 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.

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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。

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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;    }};

44 / 44 test cases passed.
Status: Accepted
Runtime: 152 ms