Java描述 LeetCode，123. Best Time to Buy and Sell Stock IV（买卖股票的时机IV）

Java描述 LeetCode，123. Best Time to Buy and Sell Stock IV（买卖股票的时机IV）

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1-1：题目

You are given an integer array prices where prices[i] is the price of a given stock on the ith day, and an integer k.

Find the maximum profit you can achieve. You may complete at most k transactions.

Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

Example 1:

Input: k = 2, prices = [2,4,1]
Output: 2
Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.

Example 2:

Input: k = 2, prices = [3,2,6,5,0,3]
Output: 7
Explanation: Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4. Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.

Constraints:

0 <= k <= 100
0 <= prices.length <= 1000
0 <= prices[i] <= 1000

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock-iv

1-2：idea

有了买卖股票3作为铺垫，这道题就比较容易了，我们只需要把各种状态通过第二个维度扩展出来即可。

public static int maxProfit(int k, int[] prices) {
        int n = prices.length;
        if (n == 0) {
            return 0;
        }
        // dp[i][j]代表第i天，
        // j代表第i天第状态，共五种。
        // - 0：当天无交易
        // - 1：第一次买入
        // - 2：第一次卖出
        // - 3：第二次买入
        // - 4：第二次卖出
        // - ....
        int[][] dp = new int[n + 1][2 * k + 1];


        // initialization
        for (int i = 0; i < dp[0].length; i++) {
            if (i % 2 == 1) {
                dp[1][i] = -prices[0];
            }
        }
        System.out.println(Arrays.deepToString(dp));
        for (int i = 2; i < n + 1; i++) {
            int price = prices[i - 1];
            dp[i][0] = dp[i - 1][0];
            for (int j = 1; j <= 2 * k; j++) {
                // j is odd
                if (j % 2 == 1) {
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - 1] - price);
                } else {
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - 1] + price);
                }
            }
//            dp[i][1] = Math.max(dp[i - 1][1], dp[i - 1][0] - price);
//            dp[i][2] = Math.max(dp[i - 1][2], dp[i - 1][1] + price);
//            dp[i][3] = Math.max(dp[i - 1][3], dp[i - 1][2] - price);
//            dp[i][4] = Math.max(dp[i - 1][4], dp[i - 1][3] + price);
        }
        System.out.println(Arrays.deepToString(dp));
        return dp[n][2 * k];
    }