0188. Best Time to Buy and Sell Stock IV (H)

Best Time to Buy and Sell Stock IV (H)

题目

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

Design an algorithm to find the maximum profit. You may complete at most k transactions.

Notice that 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 <= 10^9
• 0 <= prices.length <= 10^4
• 0 <= prices[i] <= 1000

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题意

股票买卖问题之四，允许最多k次交易(k次买入k次卖出)。

思路

解法在 0309. Best Time to Buy and Sell Stock with Cooldown 的基础上增加一个k次交易的限制。

\(hold[i][j]\)表示在第i天仍持有股票，且最多已进行了j次交易。
\(sold[i][j]\)表示在第i天未持有任何股票，且最多已进行了j次交易。

可以得到如下递推关系：

\[\begin{cases} hold[i][j]=max(sold[i-1][j-1]-prices[i],\ hold[i-1][j])\\\\ sold[i][j]=max(hold[i-1][j]+prices[i],\ sold[i-1][j]) \end{cases} \]

边界条件是：\(\forall{j\ge1},\ hold[0][j]=-prices[0]\)

需要注意的是，case可能会使坏给一个巨大的k值，导致超时。这里的一个技巧是，当k大于数组长度一半时时，问题就等同于可以进行不限次数的交易，可以直接用 0122. Best Time to Buy and Sell Stock II 中的一次遍历方法解决。

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代码实现

Java

class Solution {
    public int maxProfit(int k, int[] prices) {
        if (prices.length == 0) {
            return 0;
        }

        if (k > prices.length / 2) {
            return maxProfit(prices);
        }

        int[][] hold = new int[prices.length][k + 1];
        int[][] sold = new int[prices.length][k + 1];

        for (int i = 0; i < prices.length; i++) {
            for (int j = 1; j <= k; j++) {
                if (i == 0) {
                    hold[i][j] = -prices[i];
                } else {
                    hold[i][j] = Math.max(hold[i - 1][j], sold[i - 1][j - 1] - prices[i]);
                    sold[i][j] = Math.max(hold[i - 1][j] + prices[i], sold[i - 1][j]);
                }
            }
        }

        return sold[prices.length - 1][k];
    }

    private int maxProfit(int[] prices) {
        int profit = 0;
        for (int i = 1; i < prices.length; i++) {
            if (prices[i] > prices[i - 1]) {
                profit += prices[i] - prices[i - 1];
            }
        }
        return profit;
    }
}