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).

思路：

将vector分成两部分，用Best Time to Buy and Sell Stock I的方法求得两部分的最大值，然后求和的最大值。要求两部分的最大值需要先正面扫描一遍求得0到i的最大值，在反方向扫描一遍求i到n-1的最大值。

代码：

    int maxProfit(vector<int> &prices) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int n = prices.size();
        if(n == 0)
            return 0;
        vector<int> first(n, 0), second(n, 0);
        int min = prices[0], maxProfit = 0, max = prices[n-1];
        first[0] = 0;
        for(int i = 1; i < n; i++){
            if(prices[i] < min){
                min = prices[i];
            }
            if(prices[i] - min > maxProfit){
                maxProfit = prices[i] - min;
            }
            first[i] = maxProfit;
        }
        second[n-1] = 0;
        maxProfit = 0;
        for(int i = n-2; i >= 0; i--){
            if(prices[i] > max){
                max = prices[i];
            }
            if(max - prices[i] > maxProfit){
                maxProfit = max - prices[i];
            }
            second[i] = maxProfit;
        }
        maxProfit = first[0] + second[0];
        for(int i = 1; i < n; i++){
            if(first[i] + second[i] > maxProfit)
                maxProfit = first[i] + second[i];
        }
        return maxProfit;
    }