Best Time to Buy and Sell Stock with Transaction Fee

Best Time to Buy and Sell Stock with Transaction Fee

A.题意

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.

题目的大概意思就是你要对一只股票在n天内进行操作来获得最大利润，在n天内你可以卖出或买进股票但你手里最多只能持有这一只股票，你每次卖出股票都要交一笔手续费fee。

B.分析

很显然这是一道动态规划的问题，这n天的序列你你通过不断地卖出买进这个操作我们无法贪心，这里只能采用动态规划比较来获得最大利润的值，这里我们用两个数组sell[i]表示到第i天卖股票能得到的最大利润，hold表示第i天持有股票能获得的最大利润，于是递推式为

sell[i] = max(sell[i - 1],hold[i - 1] + prices[i] - fee);
hold[i] = max(hold[i - 1],sell[i - 1] - prices[i]);
最后的答案为sell[n]

C.代码实现

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