188. Best Time to Buy and Sell Stock IV——LeetCode

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).

Example 1:

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

Example 2:

Input: [3,2,6,5,0,3], k = 2
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.

題目大意：最多k次交易，問能夠達到的最大獲利。

思路：dp, 遞推式：

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[i, j] 表示最多i次交易時，價格爲prices[j]時的最大獲利。

核心代碼以下：

    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]);
        }
    }