[LeetCode]Best Time to Buy and Sell Stock III

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).

Have you been asked this question in an interview? Yes

思考：最多交易两次，设断点分别求前后profit，相加返回最大值，O(n²)超时。原因是重复计算，参考这里，空间换时间。哎，还是思维太局限了，不知变通。

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        int len=prices.size();
        if(len==0||len==1) return 0;
        vector<int> profit1(len);
        vector<int> profit2(len);
        int i;
        profit1[0]=0;
        int buy=prices[0];
        for(i=1;i<len;i++)
        {
            buy=min(buy,prices[i]);
            profit1[i]=max(profit1[i-1],prices[i]-buy);
        }
        profit2[len-1]=0;
        int sell=prices[len-1];
        for(i=len-2;i>=0;i--)
        {
            sell=max(sell,prices[i]);
            profit2[i]=max(profit2[i+1],sell-prices[i]);
        }
        int profit=0;
        for(i=0;i<len;i++)
        {
            if(profit<(profit1[i]+profit2[i]))
            profit=profit1[i]+profit2[i];
        }
        return profit;
    }
};