309. Best Time to Buy and Sell Stock with Cooldow
You are given an array prices where prices[i] is the price of a given stock on the ith day.
Find the maximum profit you can achieve. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times) with the following restrictions:
- After you sell your stock, you cannot buy stock on the next day (i.e., cooldown one day).
Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
题目解析：
- 三种状态：初始（冷冻期），持有，卖出
代码：
class Solution {
public int maxProfit(int[] prices) {
int[][] dp = new int[prices.length][3];
// 0：冷冻 1：持有，2：卖出
// 初始化 dp[0][1] = -prices[0];
// 状态转化
dp[0][1] = -prices[0];
for (int i = 1; i < prices.length; i++) {
dp[i][0] = Math.max(dp[i-1][0], dp[i-1][2]);
dp[i][1] = Math.max(dp[i-1][1], dp[i-1][0] - prices[i]);
dp[i][2] = Math.max(dp[i-1][2], dp[i-1][1] + prices[i]);
}
return dp[prices.length - 1][2];
}
}
714. Best Time to Buy and Sell Stock with Transact
You are given an array prices where prices[i] is the price of a given stock on the ith day, and an integer fee representing a transaction fee.
Find the maximum profit you can achieve. You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction.
Note:
- You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
- The transaction fee is only charged once for each stock purchase and sale.
代码：
class Solution {
public int maxProfit(int[] prices, int fee) {
int[][] dp = new int[2][2];
dp[0][0] = -prices[0];
for (int i = 1; i < prices.length; i++) {
dp[i%2][0] = Math.max(dp[(i-1)%2][0], dp[(i-1)%2][1] - prices[i]);
dp[i%2][1] = Math.max(dp[(i-1)%2][1], dp[(i-1)%2][0] + prices[i] - fee);
}
return dp[(prices.length-1)%2][1];
}
}
