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).

法I：把要求的东西profits设成状态。

第一次：从左往右扫描，同I一样记录下minimum value，不同的是，不仅要知道最大profit，还要记录下每个当前位置的profit（需要一个数组），为了之后与第二次扫描的结果相加。

第二次：从右往左scan，记录下maximum value, 计算profit，再加上之前第一次扫描从0到当前位置的profit，得到总的profit。

时间复杂度O(n) *2

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

        int size = prices.size();
        int maxProfit = ;
        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[] = ;
        int minValue = prices[];

       //scan from start
        for(int i = ; i< size; i++)
        {
            if(prices[i] < minValue) //save the minimum value
            {
                minValue = prices[i];
                profits[i] = profits[i-];
            }
            else //caculate the profit from minimum value to currentand compare to previous maximum profit(profits[i-1])
            {
                profits[i] = max(prices[i]-minValue,profits[i-]);
            }

        }

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

        //scan from last
        for(int i = size-; i >= ; i--)
        {
             if(prices[i] > maxValue) //save the maximum value
             {
                 maxValue = prices[i];
                 profits_from_last[i] = profits_from_last[i+];
             }
             else //caculate the profit from current to maximum value and compare to previous maximum profit(profits_from_last[i+1])
             {
                 profits_from_last[i] = max(maxValue-prices[i],profits_from_last[i+]);
             }

             int profit = profits[i] + profits_from_last[i]; //calculate profits of two transactions
             if(profit>maxProfit)
             {
                 maxProfit = profit;
             }
        }
        return maxProfit;
    }
};

Note：第二次scan不必存储每个状态，所以只需一个变量存储到目前为止第二次扫描的最大利润。

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

        int size = prices.size();
        int maxProfit = ;
        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[] = ;
        int minValue = prices[];

       //scan from start
        for(int i = ; i< size; i++)
        {
            if(prices[i] < minValue) //save the minimum value
            {
                minValue = prices[i];
                profits[i] = profits[i-];
            }
            else //caculate the profit from minimum value to currentand compare to previous maximum profit(profits[i-1])
            {
                profits[i] = max(prices[i]-minValue,profits[i-]);
            }

        }

        //initial status
        int maxSecondProfits = ;
        int maxValue = prices[size-];

        //scan from last
        for(int i = size-; i >= ; i--)
        {
             if(prices[i] > maxValue) //save the maximum value
             {
                 maxValue = prices[i];
             }
             else //caculate the profit from current to maximum value and compare to previous maximum profit(maxSecondProfits)
             {
                 maxSecondProfits = max(maxValue-prices[i],maxSecondProfits);
             }

             int profit = profits[i] + maxSecondProfits; //calculate profits of two transactions
             if(profit>maxProfit)
             {
                 maxProfit = profit;
             }
        }
        return maxProfit;
    }
};

法II：进一步提升space efficiency。扫描一次，从而不用存储第一次交易的状态。

在扫描的时候，把当前点看作两次交易都完成的点，计算当前最大利润。

然后再依次更新把当前点看作第二次买入点的最大利润，看作第一次卖出点的最大利润，第一次买入点的最大利润，这些都为了在下一个循环用来更新状态。

所以，需要用4个变量存储到当前日为止第1/2次买入/卖出的利润最大值。

时间复杂度O(n)

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int dates = prices.size();
        if(dates <= ) return ;

        int hold1 = INT_MIN, hold2 = INT_MIN;//buy stock
        int release1 = , release2 = ;//sell stock

        for(int i = ; i < dates; i++){
            release2 = max(release2, hold2+prices[i]);// The maximum if we've just sold 2nd stock so far.
            hold2 = max(hold2, release1 - prices[i]); // The maximum if we've just buy  2nd stock so far.
            release1 = max(release1, hold1+prices[i]);// The maximum if we've just sold 1nd stock so far.
            hold1 = max(hold1, -prices[i]);           // The maximum if we've just buy  1st stock so far.
        }
        return release2;
    }
};