一、题目说明
题目309. Best Time to Buy and Sell Stock with Cooldown，计算买卖股票的最大利润，卖出股票后第二天不能买入，需要冷却一天。
二、我的解答
这个题目，我没做出来。看了高手的解答，才知道用dp。
class Solution{
public:
//sold[i] = hold[i-1] + price[i];
//hold[i] = max(hold[i-1], rest[i-1] - price[i])
//rest[i] = max(rest[i-1], sold[i-1])
//最后一天max(sold，rest)
int maxProfit(vector<int>& prices){
int sold=0,rest=0,hold=INT_MIN;
for(int p: prices){
int pre_sold = sold;// sold[i-1]
sold = hold + p; //sold[i]
hold = max(hold,rest-p); // hold[i]
rest = max(rest,pre_sold);
}
return max(sold,rest);
}
};
性能如下：
Runtime: 0 ms, faster than 100.00% of C++ online submissions for Best Time to Buy and Sell Stock with Cooldown.
Memory Usage: 8.7 MB, less than 55.56% of C++ online submissions for Best Time to Buy and Sell Stock with Cooldown.
三、优化措施
dp状态方程怎么来？
每一天有3个状态：
持有hold：可以是前一天买入的继续持有；或者前一天冷却今天买入，取其最大值。
max(hold[i-1], rest[i-1] - price[i])
卖出sold: 卖出是前一天持有量 + 卖出的价格
sold[i] = hold[i-1] + price[i]
冷却rest：前一天冷却今天继续冷却，或者前一天卖出今天冷却，取其最大值。
rest[i] = max(rest[i-1], sold[i-1])