309. Best Time to Buy and Sell Stock with Cooldown

最新推荐文章于 2020-06-05 11:16:29 发布

暮雪成冰

最新推荐文章于 2020-06-05 11:16:29 发布

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本文链接：https://blog.youkuaiyun.com/u012768474/article/details/77606345

本文介绍了一种使用动态规划解决股票交易最大利润问题的方法，并通过状态机的思路清晰地阐述了状态转移过程。针对特定限制条件（如冷却期），提供了一个有效的算法实现。

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:

prices = [1, 2, 3, 0, 2]
maxProfit = 3
transactions = [buy, sell, cooldown, buy, sell]

Credits:

Special thanks to @dietpepsi for adding this problem and creating all test cases.

提示：

这道题可以用动态规划的思路解决。但是一开始想的时候总是抽象不出状态转移方程来，之后看到了一种用状态机的思路，觉得很清晰，特此拿来分享，先看如下状态转移图：

这里我们把状态分成了三个，根据每个状态的指向，我们可以得出下面的状态转移方程：

s0[i] = max(s0[i-1], s2[i-1])
s1[i] = max(s1[i-1], s0[i-1] - price[i])
s2[i] = s1[i-1] + price[i]

这样就清晰了很多。

代码：

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n=prices.size();
        if(n<=1)return 0;
        vector<int> s0(n,0);
        vector<int> s1(n,0);
        vector<int> s2(n,0);
        s1[0]=-prices[0];
        s0[0]=0;
        s2[0]=INT_MIN;
        for(int i=1;i<n;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[n-1],s2[n-1]);
    }
};