标签：算法   leetcode   

问题描述：

Say you have an array for which the i^th 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).

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;
    }

[leetcode]Best Time to Buy and Sell Stock III

标签：算法   leetcode   

原文地址：http://blog.csdn.net/chenlei0630/article/details/42274915