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).

思路：这次限定了次数 那么可以以某一天为分界点 在这一天之前最大的盈利和这一天之后的最大盈利 这两个盈利之和变为两次交易的最大盈利

#include <iostream>
#include <vector>
using namespace std;


/*
 思路：可以将两次分为历史和将来 从某一天开始 历史的最好盈利和将来的最好盈利 
 这两者的和 即为买卖两次的最佳盈利 
  
*/
int SellStockThird(vector<int>& vec)
{
	vector<int> share(vec.size(),0);
	vector<int> his(vec.size(),0);
	vector<int>	fur(vec.size(),0);
	int i,j=0;
	for(i=1;i<vec.size();i++)
		share[i] = vec[i]-vec[i-1];
	
	int cur=share[0];
	int sum = share[0];
	for(i=1;i<vec.size();i++)
	{
		if(cur < 0)
		 cur = share[i];
		else
		{
			cur+=share[i];
			his[i] = cur;
		}
			
	 	if(sum < cur)
	 	{
	 		sum = cur;
	 		his[i] = sum;
	 	}
	 	else
	 		his[i] = his[i-1];		
	}
		
	cur = share[share.size()-1];
	sum = cur;
	for(i=vec.size()-2;i>=0;i--)
	{
		if(cur < 0)
		 cur = share[i];
		else
		{
			cur+=share[i];
			fur[i] = cur;
		}
			
	 	if(sum < cur)
	 	{
	 		sum = cur;
	 		fur[i] = sum;
	 	}
	 	else
	 		fur[i] = fur[i+1];		
	}
		
	sum =0;
	
	for(i=0;i<his.size()-1;i++)
		if(sum < his[i]+fur[i+1])
			sum = his[i]+fur[i+1];
	
	return sum;
}

int main()
{
	int array[]={12,8,10,6,15,18,10};
	vector<int> vec(array,array+sizeof(array)/sizeof(int));
	cout<<SellStockThird(vec);
	return 0;
}

    public int maxProfit(int[] prices) {  
        if(prices==null || prices.length==0)  
            return 0;  
        int[] local = new int[3];  
        int[] global = new int[3];  
        for(int i=0;i<prices.length-1;i++)  
        {  
            int diff = prices[i+1]-prices[i];  
            for(int j=2;j>=1;j--)  
            {  
                local[j] = Math.max(global[j-1]+(diff>0?diff:0), local[j]+diff);  
                global[j] = Math.max(local[j],global[j]);  
            }  
        }  
        return global[2];  
    }