309. Best Time to Buy and Sell Stock with Cooldown

之前写过这题https://blog.csdn.net/weixin_43462819/article/details/83271090
但是发现竟然没有把最重要的思想写出来，这里再写一次。

之前就是用普通的dp，这里尝试另一种思路：
https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-cooldown/discuss/75928/Share-my-DP-solution-(By-State-Machine-Thinking)

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int sz = prices.size();
        if (sz < 2)
            return 0;
        vector<int> s0(sz, 0), s1(sz, 0), s2(sz, 0);
        s0[0] = 0;
        s1[0] = -prices[0];
        s2[0] = 0;
        for (int i = 1; i < sz; ++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[sz-1], s2[sz-1]);
    }
};

s0是可以买的状态；
s1是可以卖的状态；
s2是冷却的状态。

然后因为都只用了前一个状态，可以优化内存：

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int sz = prices.size();
        if (sz < 2)
            return 0;
        
        int canBuy = 0, canSell = -prices[0], rest = 0;
        for (int i = 1; i < sz; ++i) {
            int lastCanBuy = canBuy;
            canBuy = max(canBuy, rest);
            rest = canSell + prices[i];
            canSell = max(canSell, lastCanBuy-prices[i]);
        }
        return max(rest, canBuy);
    }