[leetcode-123-Best Time to Buy and Sell Stock III]

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).

开始的时候思路很朴素：以为与leetcode-121-Best Time to Buy and Sell Stock差不多

就是选取一个分割点j，在j左边与右边分别进行计算最大值，然后统计最大值之和。时间复杂度为O(n²).没有通过

最后一个数据量特别大的测试用例。

int maxProfitcore(vector<int>& prices,int start,int end)
    {//从start 到 end期间 最大profit
        if (start == end|| prices.size()==0) return 0;
        int tempProfit = 0, maxProfit = 0;
        for (int i = start+1; i <=end; i++)
        {
            tempProfit += prices[i] - prices[i - 1];
            tempProfit = max(0, tempProfit);
            maxProfit = max(maxProfit, tempProfit);
        }
        return maxProfit;
    }//超时
    int maxProfit3(vector<int>& prices)
    {
        int profit = 0;
        for (int i = 1; i < prices.size();i++)
        {
            int pro1 = maxProfitcore(prices, 0, i);
            int pro2 = maxProfitcore(prices, i, prices.size()-1);
            profit = max(profit,pro1+pro2);
        }
        return profit;
    }

后来参考大神代码：

只需扫描一趟就行，复杂度为O(n).

如下：

int maxProfitnew(vector<int>& prices)
    {
        int size = prices.size();
        if (size == 0 || size == 1) return 0;
        vector<int>profit(size,0);
        vector<int>profit1(size);        
        int local_min = prices[0];
        int local_max = prices[size - 1];
        int j = size - 2;
        int result = 0;
        profit[0] = 0;
        profit1[size - 1] = 0;
        for (int i = 1; i < size + 1 && j >= 0; i++, j--)//扫描一遍，分别计算左边和右边最大的profit
        {
            //记录i左边最大profit 利用i-1的信息 只需比较i-1时候的profit 与现在price-local_min即可
            //如果prices[i] - local_min >profit[i - 1]的话，说明需要更新最大的profit 同理 求右边最大profit也是一个道理
            profit[i] = max(profit[i - 1], prices[i] - local_min);
            local_min = min(local_min, prices[i]);
            profit1[j] = max(profit1[j + 1], local_max - prices[j]);//记录j右边最大profit
            local_max = max(local_max, prices[j]);
        }
        for (int i = 1; i < size; i++)//统计左右两边最大profit之和 取最大
        {
            result = max(result, profit[i] + profit1[i]);
        }
        return result;
    }

参考：

https://discuss.leetcode.com/topic/15177/simple-dp-8ms-solution-for-best-time-to-buy-and-sell-stock-iii

http://blog.csdn.net/fightforyourdream/article/details/14503469