题目：

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).

思路：

针对某一个时间点，分别求历史的最大盈亏与未来的最大盈亏。最终我们需要求得某一点的历史最大盈亏与未来最大盈亏最大。

代码：

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(prices.size() == 0 || prices.size() == 1)
        {
            return 0;
        }
        
        vector<int> historyProfit;
        historyProfit.assign(prices.size(),0);
        vector<int> futureProfit;
        futureProfit.assign(prices.size(),0);
        
        int min = prices[0];
        for(int i = 1; i < prices.size(); i++)
        {
            min = min<prices[i]?min:prices[i];
            int p = prices[i] - min;
            historyProfit[i] = p<historyProfit[i-1]?historyProfit[i-1]:p;
        }
        
        int max = prices[prices.size()-1];
        for(int i = prices.size()-2; i >= 0; i--)
        {
            max = max<prices[i]?prices[i]:max;
            int p = max - prices[i];
            futureProfit[i] = p<futureProfit[i+1]?futureProfit[i+1]:p;
        }
        
        int m = futureProfit[0] + historyProfit[0];
        for(int i = 1; i < prices.size(); i++)
        {
            if(futureProfit[i] + historyProfit[i] > m)
            {
                m = futureProfit[i] + historyProfit[i];
            }
        }
        return m;
    }
};