题目：

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).

算法分析：

利用通用的算法《Best Time to Buy and Sell Stock IV》 将k变为2 就ok 

AC代码：

<span style="font-family:Microsoft YaHei;font-size:12px;">public class Solution
{
    public int maxProfit(int[] prices) 
    {
        return maxProfit(2,prices);
    }
    public int maxProfit(int k, int[] prices) 
    {
       if(prices==null || prices.length==0)
            return 0;
        if(k>prices.length)//k次数大于天数时，转化为问题《Best Time to Buy and Sell Stock II》--无限次交易的情景 
        {
            if(prices==null)
                return 0;
            int res=0;
            for(int i=0;i<prices.length-1;i++)
            {
                int degit=prices[i+1]-prices[i];
                if(degit>0)
                    res+=degit;
            }
            return res;
        }
        /*
      定义维护量：
      global[i][j]:在到达第i天时最多可进行j次交易的最大利润。此为全局最优
      local[i][j]:在到达第i天时最多可进行j次交易而且最后一次交易在最后一天卖出的最大利润。此为局部最优
      定义递推式：
      global[i][j]=max(global[i-1][j],local[i][j]);即第i天没有交易，和第i天有交易
      local[i][j]=max(global[i-1][j-1]+max(diff,0),local[i-1][j]+diff)  diff=price[i]-price[i-1];
       */
        int[][] global=new int[prices.length][k+1];
        int[][] local=new int[prices.length][k+1];
        for(int i=0;i<prices.length-1;i++)
        {
            int diff=prices[i+1]-prices[i];
            for(int j=0;j<=k-1;j++)
            {
                local[i+1][j+1]=Math.max(global[i][j]+Math.max(diff,0),local[i][j+1]+diff);
                global[i+1][j+1]=Math.max(global[i][j+1],local[i+1][j+1]);
            }
        }
        return global[prices.length-1][k];
    }
}</span>