123. 买卖股票的最佳时机 III

动态规划问题，建立两个数组，存储0-i的最大增益，i-end的最大增益，然后遍历i可得到结果。

参考：

https://www.cnblogs.com/theskulls/p/5430265.html

int Max(int a, int b)
{
    return a > b ? a : b;
}

int maxProfit(vector<int>& prices) {
    if (prices.size() < 2)
        return 0;

    vector<int> lMax(prices.size(), 0);
    vector<int> rMax(prices.size()+1, 0);
    int min = prices[0];
    for (int i = 1; i < prices.size(); i++)
    {
        if (prices[i]  >= min)
        {
            lMax[i] = Max(lMax[i-1], prices[i] - min);
        }
        else
        {
            min = prices[i];
            lMax[i] = lMax[i - 1];
        }
    }
    int max = prices[prices.size()-1];
    for (int i = prices.size() - 2; i > 0; i--)
    {
        if (prices[i] <= max)
        {
            rMax[i] = Max(rMax[i + 1], max - prices[i]);
        }
        else
        {
            max = prices[i];
            rMax[i] = rMax[i+1];
        }

    }
    int result = 0;
    for (int i = 0; i < prices.size(); i++)
        result = Max(result, lMax[i] + rMax[i+1]);
    return result;
}