[LeetCode] (medium) 309. Best Time to Buy and Sell Stock with Cooldown

最新推荐文章于 2020-12-12 10:40:02 发布

转载最新推荐文章于 2020-12-12 10:40:02 发布 · 173 阅读

文章标签：

#LeetCode

LeetCode 同时被 2 个专栏收录

98 篇文章

订阅专栏

16 篇文章

订阅专栏

本文介绍了一种解决带有 cooldown 限制的股票买卖问题的算法。该算法允许进行多次交易，但每次卖出后需要等待一天才能再次购买。通过状态机 DP 方法实现，给出了一段 C++ 示例代码。

https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-cooldown/

Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times) with the following restrictions:

You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
After you sell your stock, you cannot buy stock on next day. (ie, cooldown 1 day)

Example:
Input: [1,2,3,0,2]
Output: 3 
Explanation: transactions = [buy, sell, cooldown, buy, sell]

这个状态机dp的方法实在太酷 https://www.cnblogs.com/jdneo/p/5228004.html

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if (prices.empty()) return 0;
        vector<int> S0(prices.size());
        vector<int> S1(prices.size());
        vector<int> S2(prices.size());
        
        S0[0] = 0;
        S1[0] = -prices[0];
        S2[0] = 0;
        
        for(int i = 1; i < prices.size(); ++i){
            S0[i] = max(S0[i-1], S2[i-1]);
            S1[i] = max(S0[i-1]-prices[i], S1[i-1]);
            S2[i] = S1[i-1]+prices[i];
        }
        
        return max(S0[prices.size() - 1], S2[prices.size() - 1]);
    }
};