题目要求：

最多交易两次，并且只能买卖完之后再买。

总思路：

在数组中找一个适当的点ｉ，使得i左右两边profit之和最大。

思路：

1、从左往右扫描，left[i]记录包括i元素以内的左部的maxprofit，用Best Time to Buy and Sell Stock 1可得到。

2、从右往左扫描，right[i]记录i+1到尾部元素的maxprofit，将上面的low改成big即可

3、返回max(left[i],right[i])

注意：

1、考虑好，包括i与不包括i的情况

2、考虑好头尾，left[0]肯定是0，right[n-1]和right[n-2]也是0

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