[Leetcode 123] Best Time to Buy and Sell Stock III

• Question

  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
  这是一个动态规划问题。首先最开始收入为0；
  我们用
  firstBuy=INT_MIN, firstSell=0,
  secondBuy=INT_MIN, secondSell=0;
  分别表示第一次买进的 收益（负数）， 第一次买出的 收益， 第二次买进的收益，第二次卖出的 收益。
  他们的递推公式如下：

// a表示当天的股票价格firstBuy=max(firstBuy, -a);   firstSell=max(a+firstBuy,firstSell);secondBuy=max(secondBuy,firstSell-a);secondSell=max(secondBuy+a,secondSell);

例如： prices={2,1,2,0,1}

股票价格 firstBuy firstSell secondBuy secondSell 2 -2 0 -2 0 1 -1 0 -1 0 2 -1 1 -1 1 0 0 1 1 1 1 0 1 1 2

最大收益为2.

• 代码如下：

class Solution {public:    int maxProfit(vector<int>& prices) {        int firstBuy=INT_MIN,secondBuy=INT_MIN,firstSell=0,secondSell=0;        for(auto &a：prices){            firstBuy=max(firstBuy, -a);            firstSell=max(a+firstBuy,firstSell);            secondBuy=max(secondBuy,firstSell-a);            secondSell=max(secondBuy+a,secondSell);        }        return max(secondSell,firstSell);    }};