[Leetcode] 714. Best Time to Buy and Sell Stock with Transaction Fee 解题报告-优快云博客

本文链接：https://blog.youkuaiyun.com/magicbean2/article/details/79326756

本文介绍了一种算法，用于计算在给定股票价格和交易费用的情况下，通过买卖股票所能获得的最大利润。通过定义两种状态并迭代更新，实现了时间复杂度为O(n)、空间复杂度为O(1)的高效解决方案。

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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 = 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] = 9The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.

Note:

0 < prices.length <= 50000.
0 < prices[i] < 50000.
0 <= fee < 50000.
思路：

我们定义两个状态s0和s1，其中s0表示某天拥有0支股票时的最大收益，s1表示某天拥有1支股票时的最大收益。那么在第i天，如果卖掉一支股票可以获得更大收益，则卖掉股票；否则维持原来手里没有股票时的收益。而如果买一支股票可以获得更大的收益，则买一支股票（除了支付股票费用之外还需要支付手续费）；否则维持原来手里还有一支股票。最终返回s0即可。

算法的时间复杂度是O(n)，空间复杂度是O(1)。

代码：
```
class Solution {
public:
    int maxProfit(vector<int>& prices, int fee) {
        int s0 = 0;
        int s1 = INT_MIN;
        for (int i = 0; i < prices.size(); ++i) {
            int temp = s0;
            s0 = max(s0, s1 + prices[i]);           // sell a stock at day[i]
            s1 = max(s1, temp - prices[i] - fee);   // buy a stock at day[i]
        }
        return s0;
    }
};
```