问题描述

参见上文

问题分析

选择序列Array（0..n）中任一元素Array[i]作为基点，问题转化为寻找该元素（包括该元素）之前所有元素的单次交易最大收益值MaxProfit_Array(0..i)，和该元素之后所有元素的单次交易大收益值MaxProfit_Array(i+1..n)。

寻找序列中单次交易最大收益值

• 遍历算法

    public static long calcSubMaxProfit(int[] p_samples, int startIndex, int endIndex)
    {
        long maxProfit = 0; //返回值，最大利润；
        int startSample = 0;//局部变量；买入值；
        int endSample = 0;//局部变量；卖出值；

        if (startIndex <= endIndex)
        {
            for (int i = startIndex; i < endIndex; i++) //固定买卖开始样本点
            {
                startSample = p_samples[i];
                for (int j = i+1; j <= endIndex; j++) //固定买卖结束样本点
                {
                    endSample = p_samples[j];
                    maxProfit = compareAndSwap(maxProfit, endSample - startSample);
                }
            }
        }

        return maxProfit;
    }

• 递推算法

    public static long calcSubMaxProfit_EX(int[] p_samples, int startIndex, int endIndex)
    {
        long maxProfit = 0; //返回值，最大利润；

        if (startIndex <= endIndex)
        {       
            long minSample = p_samples[startIndex];//最小买入值；
            for (int i = startIndex; i <= endIndex; i++) //卖出样本点
            {
                minSample = compareAndSwap_min(minSample, p_samples[i]);
                maxProfit = compareAndSwap(maxProfit, p_samples[i] - minSample);
            }
        }

        return maxProfit;
    }

算法实现

    /** * 计算最多两次买卖最大收益 * * @param p_samples: 样本数组； * @param len: 数组长度； * * 如果样本个数小于2，不够一次买卖操作，直接返回； * 步骤一：寻找第一次买卖节点，计算本次交易最大收益； * 步骤二：提取剩余样本数据，计算第二次交易最大收益； * 步骤三：将两次交易之和与缓存最大交易收益值比较，缓存最大值。 * * */
    public static long calcMaxProfit_EX(int[] p_samples, int len)
    {
        long maxProfit = 0; //返回值，最大利润；
        long fisrtTransProfit[] = new long[len];
        long secondTransProfit[] = new long[len];

        if (len >= 2)
        {
            for (int i = 0; i < len; i++) //固定第一次买卖开始样本点
            {
                fisrtTransProfit[i] = calcSubMaxProfit(_EX)(p_samples, 0, i);
                secondTransProfit[i] = calcSubMaxProfit(_EX)(p_samples, i+1, len-1);
                maxProfit = compareAndSwap(maxProfit, fisrtTransProfit[i]+secondTransProfit[i]);
            }
        }

        return maxProfit;
    }

性能分析

选取500个样本数据：

Array[0..499] = {10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80, 10,3,800,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80, 10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80, 10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80, 10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,300,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80,10,3,80,11,20,22,5,75,65,80};