LeetCode 714 Best Time to Buy and Sell Stock with Transaction Fee

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

Note:

0 < prices.length <= 50000.

0 < prices[i] < 50000.

0 <= fee < 50000.

思路

第i天有两种情况：持有股票及没有股票。对于持有股票，有两种操作：卖，或者不卖；对于没有股票，也有两种操作：买，或者不买。用sell[i]表示第i天没有股票（或者执行“卖”操作）所能获得的最大利润，buy[i]表示第i天持有股票（或者执行“买”操作）所能获得的最大利润，则有：

sell[i] = max(sell[i-1], buy[i-1]+prices[i]-fee)
buy[i] = max(buy[i-1], sell[i-1]-prices[i])

代码

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

后

太久没打算法题啥都不会，，，想了半天最后还是看了网上解法才略懂了，sad