Best Time to Buy and Sell Stock III

题目：

ay 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).

解析：

只能买卖两次，分别搞两个数组

动态规划

使用两个数组，forw和back。

forw[i]表示从0到i的最优买卖值（低买高卖，为正）。需要一次从左往右遍历。

back[i]表示从m-1到i的最差买卖值（高买低卖，为负）。需要一次从右往左遍历。

再对应元素相减：forw[i]-back[i]的最大值即解。

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        int m = prices.size();
        if(m == 0)
            return 0;
        
        int ret = 0;
        vector<int> forw(m, 0);
        vector<int> back(m, 0);
        
        //forward
        int curMin = prices[0];
        for(int i = 1; i < m; i  ++)
        {
            curMin = min(curMin, prices[i]);
            forw[i] = max(forw[i-1], prices[i]-curMin);
        }
        //backward
        int curMax = prices[m-1];
        for(int i = m-2; i >= 0; i --)
        {
            curMax = max(curMax, prices[i]);
            back[i] = min(back[i+1], prices[i]-curMax);
            
            ret = max(ret, forw[i]-back[i]);
        }
        
        return ret;
    }
};