188. Best Time to Buy and Sell Stock IV***

188. Best Time to Buy and Sell Stock IV***

https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/

题目描述

Say you have an array for which the i-th 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.

C++ 实现 1

推荐文章: Most consistent ways of dealing with the series of stock problems.

关于交易次数 K 进行了限制, 代码实现参考: 123. Best Time to Buy and Sell Stock III*** 中的 C++ 实现 2.

通用递推公式为:

Base cases:
T[-1][k][0] = 0, T[-1][k][1] = -Infinity
T[i][0][0] = 0, T[i][0][1] = -Infinity

Recurrence relations:
T[i][k][0] = max(T[i-1][k][0], T[i-1][k][1] + prices[i]) # sell
T[i][k][1] = max(T[i-1][k][1], T[i-1][k-1][0] - prices[i]) # buy

然而此题在实现时, 需要考虑到, 如果 K >= prices.size() / 2, 即 K 特别大的时候, (之所以 prices.size() 要除以 2, 是因为每进行一个动作, 比如 buy, 肯定要有对应的 sell 动作), 此时其实问题已经退化为, 在 prices 范围内进行无限次数的交易, 等同于 122. Best Time to Buy and Sell Stock II*.

class Solution {
public:
    int maxProfit(int K, vector<int>& prices) {
        int s0 = 0, s1 = INT32_MIN;
        // 如果 K 特别大, 超过了 prices.size() / 2 (除以 2 是因为 buy and sell 需要
        // 两个值), 此时 prices 的值不够了, 多余的 transaction 是没有意义的, 最大收益已经
        // 固定了. 其实此时相当于在 prices 上进行 infinite transactions.
        if (K >= prices.size() / 2) {
            for (int i = 0; i < prices.size(); ++ i) {
                int prev_s0 = s0;
                s0 = std::max(s0, s1 + prices[i]); // sell
                s1 = std::max(s1, prev_s0 - prices[i]); // buy 
            }
        } else {
            // 实际上, 不论 K 为多大, T 设置为 (2, prices.size() + 1) 的大小就行了
            // 但为了理解方便, 就设置为 (K + 1, prices.size() + 1) 便于推广到 K != 2 的情况.
            vector<vector<int>> T(2, vector<int>(prices.size() + 1, 0)); 
            for (int k = 0; k < K; ++ k) {
                s0 = 0, s1 = INT32_MIN;
                for (int i = 0; i < prices.size(); ++ i) {
                    s0 = std::max(s0, s1 + prices[i]); // sell
                    s1 = std::max(s1, T[k % 2][i] - prices[i]); // buy
                    T[(k + 1) % 2][i + 1] = s0; 
                }
            }
        }
        return s0;
    }
};