[leetcode] Best Time to Buy and Sell Stock III

题目：

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).

分析：把数组一分为二，正序求最大值并保存到一个数组，然后反序求最大值并保存到另一个数组，最后这两个数组相加求最终最大值即可。

代码(可参考Best Time to Buy and Sell Stock)：

    public int maxProfit(int[] prices) {
        if(prices == null || prices.length == 0)
            return 0;
        int[] result1 = new int[prices.length];//保存正序扫描的最大收益
        int[] result2 = new int[prices.length];//保存反序扫描的最大收益
        int min = prices[0], max = prices[0];
        int result = 0;
        for (int i = 1; i < prices.length; i++) {//正序扫描
            if (prices[i] > max) { // 实时计算并保存最大收益
                max = prices[i];
                if (max - min > result) {
                    result = max - min;
                }
            } else if (prices[i] < min) { // 重新规定查找区域
                min = prices[i];
                max = prices[i];
            }
            result1[i] = result;
        }
        min = prices[prices.length-1]; max = prices[prices.length-1]; result = 0;
        for (int i = prices.length-2; i >= 0; i--) {//反序扫描
            if (prices[i] < min) { // 实时计算并保存最大收益
                min = prices[i];
                if (max - min > result) {
                    result = max - min;
                }
            } else if (prices[i] > max) { // 重新规定查找区域
                min = prices[i];
                max = prices[i];
            }
            result2[i] = result;
        }
        int max_profit = 0;
        for (int i = 0; i < prices.length; i++) {//叠加求和，输出最大值
            if (result1[i] + result2[i] > max_profit) {
                max_profit = result1[i] + result2[i];
            }
        }
        return max_profit;
    }