LeetCode-188. Best Time to Buy and Sell Stock IV [C++][Java]

LeetCode-188. Best Time to Buy and Sell Stock IVhttps://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/

题目描述

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

解题思路

• buy[j] 的状态转移：

  • buy[j] = max(buy[j], sell[j-1] - prices[i])

  • 在第 i 天，可以选择不操作（保持之前的 buy[j]），或者在第 j-1 笔交易的基础上买入股票（sell[j-1] - prices[i]）。

• sell[j] 的状态转移：

  • sell[j] = max(sell[j], buy[j] + prices[i])

  • 在第 i 天，可以选择不操作（保持之前的 sell[j]），或者在当前持有股票的基础上卖出股票（buy[j] + prices[i]）。

• 时间复杂度：O(k * n)，其中 n 是天数，k 是交易次数。

• 空间复杂度：O(k)，只需要两个长度为 k+1 的数组。

【C++】

class Solution {
public:
    int maxProfit(int k, vector<int>& prices) {
        int days = prices.size();
        if(days < 2) {
            return 0;
        }
        if (k >= days) {
            int maxProfit = 0;
            for (int i = 1; i < days; ++i) {
                if (prices[i] > prices[i - 1]) {
                    maxProfit += prices[i] - prices[i - 1];
                }
            }
            return maxProfit;
        }
        vector<int> buy(k + 1, INT_MIN), sell(k + 1, 0);
        for (int i = 0; i < days; ++i) {
            for (int j = 1; j <= k; ++j) {
                buy[j] = max(buy[j], sell[j - 1] - prices[i]);
                sell[j] = max(sell[j], buy[j] + prices[i]);
            }
        }
        return sell[k];
    }
};

【Java】

class Solution {
    public int maxProfit(int k, int[] prices) {
        int days = prices.length;
        if (days < 2) {return 0;}
        if (k >= days) {
            int maxProfit = 0;
            for (int i = 1; i < days; ++i) {
                if (prices[i] > prices[i - 1]) {
                    maxProfit += prices[i] - prices[i - 1];
                }
            }
            return maxProfit;
        }
        int[] buy = new int[k + 1];
        Arrays.fill(buy, Integer.MIN_VALUE);
        int[] sell = new int[k + 1];
        for (int i = 0; i < days; ++i) {
            for (int j = 1; j <= k; ++j) {
                buy[j] = Math.max(buy[j], sell[j-1] - prices[i]);
                sell[j] = Math.max(sell[j], buy[j] + prices[i]);
            }
        }
        return sell[k];
    }
}