[leetcode] 123. Best Time to Buy and Sell Stock III 解题报告

题目链接：https://leetcode.com/problems/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 (ie, you must sell the stock before you buy again).

思路：可以交易两次，因此可以记录一下两次买和两次卖之后的最大利润．如果只能交易一次很容易看出最优解是找到后面的价格与前面的价格最大差值．而如果可以交易两次的话，第二次交易将会依赖于第一次的结果，即第二次买下一个股票剩下的利润依赖于之前第一次卖掉股票之后的利润－当前股票价格．同理第二次卖掉一支股票之后剩下的利润依赖与第二次买下一支股票之后剩下的利润＋当前股票价格．

代码如下：

class Solution {
public:
int maxProfit(vector<int>& prices) {
if(prices.size() ==0) return 0;
int buy1 = INT_MIN, buy2 = INT_MIN, sell1 = 0, sell2 =0;
for(auto price: prices)
{
buy1 = max(buy1, -price);
sell1 = max(sell1, buy1+price);
buy2 = max(buy2, sell1-price);
sell2 = max(sell2, buy2+price);
}
return sell2;
}
};

参考：https://leetcode.com/discuss/94392/clear-c-solution