188. Best Time to Buy and Sell Stock IV

class Solution {
    public int maxProfit(int k, int[] prices) {
        int n = prices.length;
        
        int[][] dp = new int[n][2 * k + 1];
        for (int i = 1; i <= k; i++) {
            dp[0][2 * i - 1] = -prices[0];
        }

        for (int i = 1; i < n; i++) {
            for (int j = 1; j < 2 * k + 1; j++) {
                dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - 1] + (((j & 1) == 0 ? 1 : -1) * prices[i]));
            }
        }

        return dp[n - 1][2 * k]; // 因为同一天可以买卖多次，所以最大值一定能转移到该状态
    }
}

Reference