leetcode—Best Time to Buy and Sell stocks III

1.题目描述

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).

2.解法分析

限定了交易次数之后题目就需要我们略微思考一下了，由于有两次交易的机会，那么我们选定一个时间点t_i，将此时间点作为两次交易的支点的话，必然有：

      t₀….t_i之内满足最佳交易原则，t_i-t_n天也满足最佳交易原则，那么这就是一个动态规划的策略了，于是有下面的代码：

 

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(prices.size() <=1)return 0;
        vector<int>::iterator iter;


        for(iter=prices.begin();iter!=prices.end()-1;++iter)
        {
            *iter = *(iter+1) - *iter;
        }

        prices.pop_back();

        vector<int>accum_forward;
        vector<int>accum_backward;

        int max = 0;
        int subMax = 0;

        for(iter=prices.begin();iter!=prices.end();++iter)
        {
            subMax += *iter;
            if(subMax > max)max=subMax;
            else
                if(subMax<0)subMax = 0;

            accum_forward.push_back(max);
        }

        vector<int>::reverse_iterator riter;

        max =0 ;
        subMax = 0;
        for(riter=prices.rbegin();riter!=prices.rend();++riter)
        {
            subMax +=*riter;
            if(subMax >max)max = subMax;
            else
                if(subMax<0)subMax=0;

            accum_backward.push_back(max);
        }

        max =0;
        int len = accum_forward.size();
        for(int i=0;i<len-1;++i)
        {
            if((accum_forward[i]+accum_backward[len-i-2])>max)
                max = accum_forward[i]+accum_backward[len-i-2];
        }

        return max>accum_forward[len-1]?max:accum_forward[len-1];

    }
};

 

ps:做完题之后提交，发现老是AC不了，有个case总是解决不了，本来以为是自己代码写得有问题，检查了半天没发现错误，于是开始看别人怎么写，结果发现别人AC的代码也过不了，猜想可能系统还是做得不完善，应该是后台的线程相互干扰了，过了一段时间果然同样的代码又可以AC了。在这段过程中，看别人写的代码，发现了一个比我简洁一些的写法，虽然我么你的复杂度是一样的，但是此君代码量比我的小一点，以后学习学习，另外，一直不知道vector还可以预先分配大小，从这个代码里面也看到了，算是有所收获。附代码如下：

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        // null check
        int len = prices.size();
        if (len==0) return 0;

        vector<int> historyProfit;
        vector<int> futureProfit;
        historyProfit.assign(len,0);
        futureProfit.assign(len,0);
        int valley = prices[0];
        int peak = prices[len-1];
        int maxProfit = 0;

        // forward, calculate max profit until this time
        for (int i = 0; i<len; ++i)
        {
            valley = min(valley,prices[i]);
            if(i>0)
            {
                historyProfit[i]=max(historyProfit[i-1],prices[i]-valley);
            }
        }

        // backward, calculate max profit from now, and the sum with history
        for (int i = len-1; i>=0; --i)
        {
            peak = max(peak, prices[i]);
            if (i<len-1)
            {
                futureProfit[i]=max(futureProfit[i+1],peak-prices[i]);
            }
            maxProfit = max(maxProfit,historyProfit[i]+futureProfit[i]);
        }
        return maxProfit;
    }

};