309. Best Time to Buy and Sell Stock with Cooldown

Problem

Say you have an array for which the ithelement 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:

prices = [1, 2, 3, 0, 2]

maxProfit = 3

transactions = [buy, sell, cooldown, buy, sell]

Solution

讨论区的解答



There are three states, according to the action that you can take.

Hence, from there, you can now the profit at a state at time i as:

s0[i] = max(s0[i - 1], s2[i - 1]); // Stay at s0, or rest from s2

s1[i] = max(s1[i - 1], s0[i - 1] - prices[i]); // Stay at s1, or buyfrom s0

s2[i] = s1[i - 1] + prices[i]; // Only one way from s1

Then, you just find the maximum of s0
and s2
, since they will be the maximum profit we need (No one can buy stock and left with more profit that sell right :) )

Define base case:

s0[0] = 0; // At the start, you don’t have any stock if you just rest

s1[0] = -prices[0]; // After buy, you should have -prices[0] profit.Be positive!

s2[0] = INT_MIN; // Lower base case

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

s0[0] = 0;
s1[0] = -prices[0];
s2[0] = INT_MIN;
for(int i = 1;i<n;++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]);

}
};