Leetcode 算法习题 第十四周

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.

题目大意

给定每天的价格，决定买入卖出日使得收益最大，输出最大收益

我的解答

class Solution {
public:
    int maxProfit(vector<int>& prices, int fee) {
        int days = prices.size();
        vector<int> buy(days+1,0);
        vector<int> sell(days+1,0);
        buy[0] = -199999;
        for(int i = 1; i <= days; i++){
            buy[i] = max(buy[i-1],sell[i-1]-fee-prices[i-1]);//不买（维持前一天的收益累积）或买（前一天已有的收益-fee-今天的价格）
            sell[i] = max(sell[i-1],buy[i-1] + prices[i-1]);//不卖或卖
        }
        return sell[days];
    }
};