leetcode练习 714. Best Time to Buy and Sell Stock with Transaction Fee
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动态规划感觉很重要,很锻炼思维。
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 = 2Output: 8Explanation: 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.
思路如下:
定义:
hold:买了石头后手头的钱
sell:卖了石头后手头的钱
状态转移:
当天卖了石头后手头的钱=max(前一天卖了石头后手头的钱,今天做了交易之后手头的钱)
当天买了石头后手头的钱=max(前一天买了石头后手头的钱,今天用手头的钱买了石头后手头的钱)
代码:
class Solution {public: int maxProfit(vector<int>& prices, int fee) { int hold = -prices[0]; int sell = 0; for (int i = 0; i < prices.size(); i++) { sell = max(sell, hold + prices[i] - fee); hold = max(hold, sell - prices[i]); } return sell; }};
看似很短,更重要的是思路
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