Problem Description:

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).

按照题意，要求根据每天的股票价格，交易两次得到最大的利润。直接的想法就是利用分治法，从前往后循环依次将数组分为前后两个股票价格序列，分别得到最大的利润，将两者相加得到两次交易的最大利润，时间复杂度O(n^2)，提交果然超时了。

代码如下：

class Solution {
public:

    int profit(vector<int> &vec,int beg,int last)
    {
        int res=0;
        if(beg==last)
            return res;
        int min=vec[beg];
        for(int i=beg+1;i<=last;++i)
        {
            if(vec[i]>min)
            {
                int temp=vec[i]-min;
                if(temp>res)
                    res=temp;
            }
            else
                min=vec[i];
        }
        return res;
    }

    int maxProfit(vector<int> &prices) {
        int res=0;
        int n=prices.size();
        if(n<=1)
            return res;

        for(int i=1;i<n-1;i++)
        {
            int res1=profit(prices,0,i);
            int res2=profit(prices,i+1,n-1);
            if((res1+res2)>res)
                res=res1+res2;
        }

        return res;
    }
};

具体代码如下：

class Solution {
public:

    int maxProfit(vector<int> &prices) {
        int res=0;
        int n=prices.size();
        if(n<=1)
            return res;
        vector<int> front(n,0);
        vector<int> back(n,0);
        int minp=prices[0];
        int maxp=prices[n-1];
        for(int i=1;i<n;i++)
        {
            if(prices[i]>minp)
                front[i]=prices[i]-minp;
            else
                minp=prices[i];
        }
        for(int i=n-2;i>=0;i--)
        {
            if(prices[i]>maxp)
                maxp=prices[i];
                
            back[i]=max(maxp-prices[i],back[i+1]);
            if(front[i]+back[i]>res)
                res=front[i]+back[i];
                
        }

        return res;
    }
};