LEETCODE : best-time-to-buy-and-sell-stock

 题目描述

假设你有一个数组，其中第i个元素是某只股票在第i天的价格。
设计一个算法来求最大的利润。你最多可以进行两次交易。
注意:
你不能同时进行多个交易(即，你必须在再次购买之前出售之前买的股票)。
 

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 two transactions.

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

解：核心思想是每次交易完 赔赚的钱到底是多少，或者说钱包里还剩多少钱。

 

class Solution {
public:

    int maxProfit(vector& prices) {
        // write code here
        int buy1 = INT_MIN,buy2 = INT_MIN;
        int sell1 = 0,sell2 =0;
        for(auto pri:prices){
            buy1 = max(buy1, -pri);
            sell1 = max(sell1, buy1+pri);
            buy2 = max(buy2, sell1-pri);
            sell2 = max(sell2, buy2+pri);
        }
        return sell2;
    }
};

图中的数字就是经过一次交易（买/卖）之后剩余的钱。