123. Best Time to Buy and Sell Stock III (Array; DP)

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：把要求的东西profits设成状态。 从左往右，从右往左两次scan赋值状态，然后再扫描一次状态，所以时间复杂度O(n) *3

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        if(prices.empty()) return 0;

        int size = prices.size();
        int maxProfit = 0;
        int* profits = new int [size]; //表示从0到i的最大利润
        int* profits_from_last = new int [size]; //表示从i到size-1的最大利润

        //initial status
        profits[0] = 0;
        int minValue = prices[0];

       //scan from start
        for(int i = 1; i< size; i++)
        {
            if(prices[i] < minValue)
            {
                minValue = prices[i];
                profits[i] = profits[i-1];
            }
            else
            {
                profits[i] = max(prices[i]-minValue,profits[i-1]);
            }

        }

        //initial status
        profits_from_last[size-1] = 0;
        int maxValue = prices[size-1];

        //scan from last
        for(int i = size-2; i >=0 ; i--)
        {
             if(prices[i] > maxValue)
             {
                 maxValue = prices[i];
                 profits_from_last[i] = profits_from_last[i+1];
             }
             else
             {
                 profits_from_last[i] = max(maxValue-prices[i],profits_from_last[i+1]);
             }

             int profit = profits[i] + profits_from_last[i];
             if(profit>maxProfit)
             {
                 maxProfit = profit;
             }
        }
        return maxProfit;
    }
};