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[Leetcode]188. Best Time to Buy and Sell Stock IV (C++)

题目描述

题目链接:188. Best Time to Buy and Sell Stock IV

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).

例子

例子 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.

例子 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.

解题思路

方法一

这道题是买卖股票的第四题,具体来说是第三题[Leetcode]123. Best Time to Buy and Sell Stock III (C++) 的一般化版本。这里我还是沿用了第三题的思路,记录每一次买入和卖出时的利润,通过和前一次卖出结果进行比较和更新,是一种动态规划的思路。除此之外,当最大买卖次数过多(超过prices 的总量时,可以约束 k = prices.size() 来降低所需时间),代码如下:

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#include <vector>
#include<bits/stdc++.h>

class Solution {
public:
int maxProfit(int k, std::vector<int>& prices) {
if (k == 0 || prices.size() <= 1) return 0;
if (k > prices.size()) k = prices.size() / 2;
std::vector<int> kth_buy(k, INT_MIN);
std::vector<int> kth_sell(k, 0);

for (int currentPrice: prices) {
for (int i = k - 1; i > 0; i--) {
kth_sell[i] = std::max(kth_sell[i], currentPrice + kth_buy[i]);
kth_buy[i] = std::max(kth_buy[i], kth_sell[i - 1] - currentPrice);
}
kth_sell[0] = std::max(kth_sell[0], currentPrice + kth_buy[0]);
kth_buy[0] = std::max(kth_buy[0], -currentPrice);
}

return kth_sell.back() < 0? 0 : kth_sell.back();
}
};
  • 时间复杂度: O(kn)
  • 空间复杂度: O(k)

方法二

方法一应该不是最优解,有空再学习一下把最优解补上。