Question 1: 309. The best time to buy and sell stocks includes a freezing period - LeetCode
Given an array of integers prices, where the th represents the stock price on the th day. prices[i]i
Design an algorithm to calculate the maximum profit. You can complete as many transactions as possible (buy and sell a stock multiple times) subject to the following constraints:
- After selling the stock, you cannot buy the stock the next day (that is, the freezing period is 1 day).
Note: You cannot participate in multiple transactions at the same time (you must sell the previous stock before buying again)
Idea: There is an extra cooling-off period for this question, which is more troublesome. The key lies in the description of the state. The code is as follows:
class Solution {
public:
int maxProfit(vector<int>& prices) {
int n = prices.size();
if (n == 0) return 0;
vector<vector<int>> dp(n, vector<int>(4, 0));
dp[0][0] -= prices[0]; // 持股票
for (int i = 1; i < n; i++) {
dp[i][0] = max(dp[i - 1][0], max(dp[i - 1][3] - prices[i], dp[i - 1][1] - prices[i]));
dp[i][1] = max(dp[i - 1][1], dp[i - 1][3]);
dp[i][2] = dp[i - 1][0] + prices[i];
dp[i][3] = dp[i - 1][2];
}
return max(dp[n - 1][3], max(dp[n - 1][1], dp[n - 1][2]));
}
};
Question 2: 714. The best time to buy and sell stocks includes handling fees - LeetCode
Given an array of integers prices, which prices[i]represent i the stock prices on the first day; the integers fee represent the handling fees for trading stocks.
You can complete transactions an unlimited number of times, but you need to pay a transaction fee for each transaction. If you have already bought a stock, you cannot continue to buy another stock until you sell it.
Returns the maximum value of profit.
Note: A transaction here refers to the entire process of buying, holding and selling stocks, and you only need to pay a handling fee for each transaction.
Idea: There are more handling fees for this question, so we should think more about how to make a reasonable description. The specific code is as follows:
class Solution {
public:
int maxProfit(vector<int>& prices, int fee) {
int n = prices.size();
vector<vector<int>> dp(n, vector<int>(2, 0));
dp[0][0] -= prices[0]; // 持股票
for (int i = 1; i < n; i++) {
dp[i][0] = max(dp[i - 1][0], dp[i - 1][1] - prices[i]);
dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] + prices[i] - fee);
}
return max(dp[n - 1][0], dp[n - 1][1]);
}
};