[leetcode]Best Time to Buy and Sell Stock III

问题描述：

Say you have an array for which the i^th element is the price of a given stock on dayi.

Design an algorithm to find the maximum profit. You may complete at mosttwo transactions.

Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

基本思想：

动态规划方法

'''

// f[k, ii] represents the max profit up until prices[ii] (Note: NOT ending with prices[ii]) using at most k transactions.         // f[k, ii] = max(f[k, ii-1], prices[ii] - prices[jj] + f[k-1, jj]) { jj in range of [0, ii-1] }        //          = max(f[k, ii-1], prices[ii] + max(f[k-1, jj] - prices[jj]))        // f[0, ii] = 0; 0 times transation makes 0 profit        // f[k, 0] = 0; if there is only one price data point you can't make any money no matter how many times you can trade        if (prices.size() <= 1) return 0;

'''

代码：

int maxProfit(vector<int> &prices) {  //C++        int size = prices.size();        if(size <=1)            return 0;                    int transNum = 2;        vector <vector<int> > profit(transNum+1,vector<int>(size,0));                    int maxprofit = 0;        for(int i=1; i<=transNum; i++)        {            int tempmax = profit[i-1][0]-prices[0];            for(int j = 1; j< size; j++)            {                profit[i][j] = max(profit[i][j-1],prices[j]+tempmax);                tempmax = max(tempmax,profit[i-1][j]-prices[j]);                maxprofit = max(maxprofit,profit[i][j]);            }        }        return maxprofit;    }