leetcode 714. Best Time to Buy and Sell Stock with Transaction Fee【贪心？？水题】

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本文介绍了一个算法问题，即如何在给定股票价格和交易费用的情况下，通过买卖操作获得最大利润。文章提供了一段C++代码实现，利用动态规划的方法解决该问题。

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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] = 1Selling at prices[3] = 8Buying at prices[4] = 4Selling 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.

本来以为是dp单调队列优化，被题意误导了，小哥哥说是自己阿里面试题，都把dp公式给我了，自己初始状态想了半天。

太太太菜了QAQ

class Solution {
public:
    int maxProfit(vector<int>& prices, int fee) {
        int sold[50005],hold[50005];
        memset(sold,0,sizeof(sold));
        memset(hold,0,sizeof(hold));
        sold[0]=0;
        hold[0]=-prices[0];
        int n=prices.size();
        for(int i=1;i<n;i++)
        {
            sold[i]=max(sold[i-1],hold[i-1]+prices[i]-fee);
            hold[i]=max(hold[i-1],sold[i-1]-prices[i]);
        }
        return sold[n-1];
    }
};