Leetcode 309 best time to buy and sell shares with the freezing period (dynamic programming)

Is to find a state transition equation, the problem can be solved

But the state transition equation easy to find ah

Analysis Title:

Four states every day: buy, sell, freezing period, do nothing

The arrangement follows the daily status: buy ... buy ... sell ... sell cold cold ... ... which represents the day doing nothing, there may be multiple

Because some day doing nothing, no way to specify a certain day in the end was kind of operation, so the state as "the last operation of the day and before"

And the i-th day before the final operation is to buy the biggest gains obtained: sell [i] 

Similarly defined buy [i], cool [i]

State transition equation:

sell[i]=buy[i-1]+price[i]
buy[i]=max(cool[i-1]-prices[i],buy[i])
cool[i]=max(sell[i-1],cool[i-1])

Code:

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int len = prices.size();
        if (len == 0)return 0;
        vector<int> sell(len), buy(len), cool(len);
        sell[0] = 0;
        buy[0] = -1*prices[0];
        cool[0] = 0;
        for (int i = 1;i < len;i++) {
            sell[i] = buy[i - 1] + prices[i];
            buy[i] = max(cool[i - 1] - prices[i], buy[i - 1]);
            cool[i] = max(sell[i - 1], cool[i - 1]);
        }
        int ans = max(sell[len-1], buy[len-1]);
        years = max (age, cool [len - 1 ]);
        return years;
    }
};