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).

想不出来，网上搜了一个算法，从头算一遍，然后在从尾算一遍当前最大值，然后从不同的i分割，前后加起来最大的情况。另外还看到了一个求最大k次交易的一个通用算法，是根据一篇论文写出来的，看来学术确实能用到平常的编程里来，其实想想现在的二分查找等算法不都是以前的论文么。

int maxProfitIII(vector<int> &prices)
{
int len = prices.size();
if (len == 0) return 0;

vector<int> history(len, 0);
vector<int> future(len, 0);

int low = prices[0];
for (int i = 1; i < len; i++)
{
history[i] = max(history[i-1], prices[i] - low);
low = min(low, prices[i]);
}

int high = prices[len-1];
for (int i = len-2; i >= 0; i--)
{
future[i] = max(future[i+1], high - prices[i]);
high = max(high, prices[i]);
}

int maxProfit = 0; // 网上说第一次的卖和第二次的买可以在同一次，所以直接都是i相加就可以了
for (int i = 0; i < len; i++)
{
maxProfit = max(maxProfit, history[i] + future[i]);
}

return maxProfit;
}