111_leetcode_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).

1:特殊情況；2:從前到後和從後到前遍曆兩次數字；3:在遍曆的過程中，儲存以當前索引為結束點或者開始點的只進行一次買賣的最大收益；4:注意邊界結束條件

        int maxProfit(vector<int> &prices)    {        if(prices.size() <= 1)        {            return 0;        }                int size = (int)prices.size();        vector<int> leftProfit(size, 0);        leftProfit[0] = 0;        int minValue = prices[0];                int result = 0;                for(int i = 1; i < size; i++)        {            if(prices[i] > minValue)            {                leftProfit[i] = (prices[i] - minValue > leftProfit[i-1] ? prices[i] - minValue : leftProfit[i-1]);            }            else            {                minValue = prices[i];                leftProfit[i] = leftProfit[i-1];            }        }                result = leftProfit[size-1] > leftProfit[size-2] ? leftProfit[size-2] : leftProfit[size-1];        int maxValue = prices[size -1];        int rightMaxProfit = 0;                for(int i = size - 2; i >= 0; i--)        {            if(prices[i] < maxValue)            {                rightMaxProfit = (rightMaxProfit > maxValue - prices[i] ? rightMaxProfit : maxValue - prices[i]);            }            else            {                maxValue = prices[i];            }                        if(i == 0)            {                result = (result > rightMaxProfit ? result : rightMaxProfit);            }            else            {                result = (result > rightMaxProfit + leftProfit[i-1] ? result : rightMaxProfit + leftProfit[i-1]);            }        }                return result;    }