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  • [leetcode] 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:

    • 0 < prices.length <= 50000.
    • 0 < prices[i] < 50000.
    • 0 <= fee < 50000.

    用maxProfit[i]记录前i个数据的最大收益。

    用cashWhenHoldOne[i]表示对于前i个数据,在手中有且只有一个stock的情况下,手中现金的最大值。

    则:

    maxProfit[i] = max(maxProfit[i-1], cashWhenHoldOne[i-1] + prices[i] - fee);
    cashWhenHoldOne[i] = max(cashWhenHoldOne[i-1], maxProfit[i-1] - prices[i]);

    我的代码如下:

    class Solution {
    public:
        int maxProfit(vector<int>& prices, int fee) {
            int len = prices.size();
            vector<int> maxProfit(len), cashWhenHoldOne(len);
            maxProfit[0] = 0, cashWhenHoldOne[0] = -prices[0];
            for (int i = 1; i < len; i++) {
                maxProfit[i] = max(maxProfit[i-1], cashWhenHoldOne[i-1] + prices[i] - fee);
                cashWhenHoldOne[i] = max(cashWhenHoldOne[i-1], maxProfit[i-1] - prices[i]);
            }
            return maxProfit[len-1];
        }
    };
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  • 原文地址:https://www.cnblogs.com/zmj97/p/7892727.html
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