see-123. Best Time to Buy and Sell Stock III

本文链接：https://blog.csdn.net/leetcodecl/article/details/85240103

本文介绍了一种算法，旨在通过最多两笔交易获得股票价格数组中可能的最大利润。文章通过三个实例详细解释了算法的工作原理，强调了不能同时进行多次交易的规则，并提供了实现该算法的代码示例。

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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).

Example 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.

Example 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.

Example 3:

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

//无论买卖几次，始终保证当前手里收益最大
 int maxProfit(vector<int>& prices) {
        if(prices.size()<2)
            return 0;
        
        int FirstBuy=INT_MIN,FirstSell=INT_MIN;
        int secondBuy=INT_MIN,secondSell=INT_MIN;       
        for(int i=0;i<prices.size();i++)
        {
            FirstBuy=max(FirstBuy,-prices[i]); // 买完第一次，手里的利益
            FirstSell=max(FirstSell, prices[i]+FirstBuy);//卖完第一次，手里的利益
            secondBuy=max(secondBuy,-prices[i]+FirstSell);// 买完第二次，手里的利益
            secondSell=max(secondSell,prices[i]+secondBuy);//卖完第二次，手里的利益
        }
        return secondSell;
    }