题目描述

题目链接：123. Best Time to Buy and Sell Stock III

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 two transactions.

Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

例子

例子 1

Input: [3,3,5,0,0,3,1,4] Output: 6 Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3. Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.

例子 2

Input: [1,2,3,4,5] Output: 4 Explanation:Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4. Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are engaging multiple transactions at the same time. You must sell before buying again.

例子 3

Input: [7,6,4,3,1] Output: 0 Explanation: In this case, no transaction is done, i.e. max profit = 0.

解题思路

这道题是买卖股票的第三题，要求最多只能交易两次。参考了一下网上的做法，有一种我觉得比较好理解的方法是，维护四个变量，分别是：

• FirstBuy : 第一次买入时的利润（应该是负值）
• FirstSell：第一次卖出时的利润
• SecondBuy：第二次买入时（此时第一次已卖出）的利润（可正可负）
• SecondSell：第二次卖出时的利润

遍历一次数组，依序更新：

• SecondSell 比较 SecondSell + prices[i]
• SecondBuy 比较 FirstSell - prices[i] (买入，所以利润相减)
• FirstSell 比较 FirstBuy + prices[i]
• FirstBuy 比较 -prices[i]

代码如下：

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

class Solution {
public:
    int maxProfit(std::vector<int>& prices) {
        if (prices.size() <= 1) return 0;
        int firstBuy = INT_MIN; //profit after first buy in, should be negative
        int firstSell = 0; // profit after fisrt sell, should be positive
        int secondBuy = INT_MIN; //profit after second buy in, might be negative or positive
        int secondSell = 0; // profit after second sell, final answer

        for (int currentPrice: prices) {
            secondSell = std::max(secondSell, currentPrice + secondBuy);
            secondBuy = std::max(secondBuy, firstSell - currentPrice);
            firstSell = std::max(firstSell, currentPrice + firstBuy);
            firstBuy = std::max(firstBuy, -currentPrice);

        return secondSell < 0 ? 0 : secondSell;

    }
};

• 时间复杂度：O(n)
• 空间复杂度：O(1)