[LeetCode][309]best-time-to-buy-and-sell-stock-with-cooldown

佚名 / 2023-08-29 / 原文

Content

You are given an array prices where prices[i] is the price of a given stock on the i^th 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).

Example 1:

Input: prices = [1,2,3,0,2]
Output: 3
Explanation: transactions = [buy, sell, cooldown, buy, sell]

Example 2:

Input: prices = [1]
Output: 0

Constraints:

1 <= prices.length <= 5000
0 <= prices[i] <= 1000

Solution

1. 动态规划

Java

class Solution {
    public int maxProfit(int[] prices) {
        // 1 <= prices.length <= 5000
        // 0 <= prices[i] <= 1000
        int n = prices.length;
        // dp[i]表示第i天结束后累计最大收益
        // dp[i][0]表示第i天结束后，持有股票累计最大收益
        // dp[i][1]表示第i天结束后，今天刚卖出股票累计最大收益
        // dp[i][2]表示第i天结束后，未持有股票并且今天没有任何操作累计最大收益
        int[][] dp = new int[n][3];
        dp[0][0] = -prices[0];
        dp[0][1] = 0;
        dp[0][2] = 0;
        for (int i = 1; i < n; i++) {
            dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][2] - prices[i]);
            dp[i][1] = dp[i - 1][0] + prices[i];
            dp[i][2] = Math.max(dp[i - 1][1], dp[i - 1][2]);
        }
        return Math.max(dp[n - 1][1], dp[n - 1][2]);
    }
}