leetcode Best Time to Buy and Sell Stock III

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题目要求之多两次事务得到的最大收益。

较为自然的联想到以时间 i 为分界点，求出 0- - -i 与 i+1- - -end之间的最大效益，分别比较大小以及两者之和的大小即可获得最大利润。

代码（超时）

class Solution {public:    int maxProfit(vector<int> &prices) {        maxNum = 0;        if(prices.size()==0||prices.size()==1)            return 0;                for(int i = prices.size()-1; i >= 0; --i)        {            int sum1 = getMax(0, i-1, prices);            int sum2 = getMax(i, prices.size()-1, prices);            int sum3 = sum1+sum2;            maxNum = max(sum3, max(sum1, sum2));                    }                return maxNum;    }            int getMax(int start, int end, vector<int> &prices)    {        if(start==end||start>end||start<0||end<0)          return 0;        int maxSold = prices[end];        int maxNum = 0;        for(int i = end; i >= 0; --i)        {            maxSold = max(prices[i], maxSold);            maxNum = max(maxNum, maxSold - prices[i]);                    }                return maxNum;            }           private:    int maxNum;};

上述算法时间复杂度为O(n^2)

参考网上资料可以优化代码（Accepted）

class Solution {public:    int maxProfit(vector<int> &prices) {        maxNum = 0;        int len = prices.size();        if(len<=1)            return 0;        int maxFromHead[len];                int minBought = prices[0];        int maxPrice = 0;                for(int i = 0; i < prices.size(); ++i)        {           minBought = min(minBought, prices[i]);           maxPrice = max(maxPrice, prices[i]-minBought);           maxFromHead[i] = maxPrice;                    }                        int maxSold = prices[len-1];        maxPrice = 0;        maxNum = maxFromHead[len-1];        for(int i = len-1; i >= 0; --i)        {            maxSold = max(maxSold, prices[i]);            maxPrice = max(maxPrice, maxSold - prices[i]);            maxNum = max(maxNum, maxPrice+maxFromHead[i-1]);        }                return maxNum;    }              private:    int maxNum;};

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