188. Best Time to Buy and Sell Stock IV

Say you have an array for which the ith element is the price of a given stock on day i.
Design an algorithm to find the maximum profit. You may complete at most k transactions.
Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

Solution：DP

思路：

Screen Shot 2017-11-27 at 22.06.00.png

Time Complexity: O(mn) Space Complexity: O(mn)

Solution Code：

class Solution {
    /**
     * dp[i, j] represents the max profit up until prices[j] using at most i transactions. 
     * dp[i, j] = max(dp[i, j-1], prices[j] - prices[jj] + dp[i-1, jj]) { jj in range of [0, j-1] }
     *          = max(dp[i, j-1], prices[j] + max(dp[i-1, jj] - prices[jj]))
     * dp[0, j] = 0; 0 transactions makes 0 profit
     * dp[i, 0] = 0; if there is only one price data point you can't make any transaction.
     */

    public int maxProfit(int k, int[] prices) {
        int n = prices.length;
        if (n <= 1) return 0;

        //if k >= n / 2, then you can make maximum number of transactions.
        if (k >= n / 2) {
            int maxPro = 0;
            for (int i = 1; i < n; i++) {
                if (prices[i] > prices[i-1])
                    maxPro += prices[i] - prices[i-1];
            }
            return maxPro;
        }

        int[][] dp = new int[k + 1][n];
        for (int i = 1; i <= k; i++) {
            int localMax = dp[i - 1][0] - prices[0];
            for (int j = 1; j < n; j++) {
                dp[i][j] = Math.max(dp[i][j - 1],  prices[j] + localMax);
                localMax = Math.max(localMax, dp[i-1][j] - prices[j]);
            }
        }
        return dp[k][n - 1];
    }
}