[WXM] LeetCode 714. Best Time to Buy and Sell Stock with Transaction Fee C++

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

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:

1. 0 < prices.length <= 50000.
2. 0 < prices[i] < 50000.
3. 0 <= fee < 50000.

Approach

1. 这道题很有意思，题目大意是给你股票的行情和转让费，问你怎么让收益最大化。这道题归类到数学和动态规划里，所以我们用动态规划的方式解题，但是因为这也涉及到数学，所以你要计算得出结论，怎样才能收益，然后才能借用动态规划的模型去解。结论就是当你卖股票盈余的钱（也就是cash = max(cash, hold + prices[i] - fee)）买得起下只股票，才有可能继续盈余（也就是hold = max(hold, cash - prices[i])）。

Code

class Solution {
public:
    int maxProfit(vector<int>& prices, int fee) {
        int cash = 0, hold = -prices[0];
        for (int i = 1; i < prices.size(); i++) {
            cash = max(cash, hold + prices[i] - fee);
            hold = max(hold, cash - prices[i]);
        }
        return cash;
    }
};