309. Best Time to Buy and Sell Stock with Cooldown

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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:

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> 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

题意：给出股票的每天的价格，求买卖股票的最大收益。有两个约束条件：

1、 卖股票之前必要有股票在手.

2、卖完股票的时候必须隔一天才能再买（cooldown）。

状态机

Code:

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//
// Created by cpz on 2017/11/28.
//
#include "head.h"

class Solution {
public:
    int maxProfit(vector<int> &prices)
    {
        int len = prices.size();
        if (len <= 1)
            return 0;
        vector<int> s0;
        vector<int> s1;
        vector<int> s2;

        s0.push_back(0);
        s1.push_back(-prices[0]);
        s2.push_back(INT_MIN);

        for (int i = 1; i < len; ++i)
        {
            s0.push_back(max(s0[i - 1], s2[i - 1]));
            s1.push_back(max(s0[i - 1] - prices[i], s1[i - 1]));
            s2.push_back(s1[i - 1] + prices[i]);
        }
        return max(s0[len - 1], s2[len - 1]);
    }
};