LeetCode--Best Time to Buy and Sell Stock with Cooldown

Say you have an array for which the i^th element is the price of a given stock on dayi.

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]

本题的题意就是每天股票都有股价，可以从买入，卖出和不进行操作这三个选项中做出选择，问最后一天过后最大的利润是多少。

根据买入，卖出和不进行操作这三个选项可以得出图示的三种状态，s0代表可前一天之后不操作得到的状态，s1代表前一天之后进行买入操作或者不操作得到的状态，s2代表前一天之后进行卖出操作得到的状态。

那么状态转移方程就为：

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]

代码如下：

class Solution {public:    int maxProfit(vector<int>& prices){        if (prices.size() <= 1) return 0;        int s0[10001];        int s1[10001];        int s2[10001];        memset(s0, 0, sizeof(s0));        memset(s1, 0, sizeof(s1));        memset(s2, 0, sizeof(s2));        s1[0] = -prices[0];        s0[0] = 0;        s2[0] = -999999;        for (int i = 1; i < prices.size(); i++) {            s0[i] = max(s0[i - 1], s2[i - 1]);            s1[i] = max(s1[i - 1], s0[i - 1] - prices[i]);            s2[i] = s1[i - 1] + prices[i];        }        return max(s0[prices.size() - 1], s2[prices.size() - 1]);    }};

这里要注意的是最后一定是从s0状态和s2状态中选最大值，因为s1是经过前一个状态买股票得到的，那么它的值一定会小于前一个状态，所以s1[prices.size()-1]一定会小于最大值。