java编程面试——风口的猪-中国牛市

题目描述

风口之下，猪都能飞。当今中国股市牛市，真可谓“错过等七年”。 给你一个回顾历史的机会，已知一支股票连续n天的价格走势，以长度为n的整数数组表示，数组中第i个元素（prices[i]）代表该股票第i天的股价。 假设你一开始没有股票，但有至多两次买入1股而后卖出1股的机会，并且买入前一定要先保证手上没有股票。若两次交易机会都放弃，收益为0。 设计算法，计算你能获得的最大收益。 输入数值范围：2<=n<=100,0<=prices[i]<=100

输入例子:

3,8,5,1,7,8

输出例子:

12

/** * 计算你能获得的最大收益 * * @param prices Prices[i]即第i天的股价 * @return 整型 */

    public int
calculateMax(int[] prices) {

          int max=0;

          int a1=0;  

        // 只卖一次

               for(int i=0;i<=(prices.length-2);i++)

            {

                    a1=0;

                  for(int j=i+1;j<=(prices.length-1);j++)

                           {

                               if(prices[j]>a1)

                                   {

                                  a1= prices[j];

                               }

                       }

                    if(max<(a1-prices[i]))

                               {

                               max=a1-prices[i];

                           }

             }

                  //可以卖出两次，确定两个买进点的值，遍历两段间最大值，然后作差和保留值比较

        if(prices.length>=4)

            {

                        for(int i=0;i<=(prices.length-4);i++)

                {

                       for(int j=i+2;j<=(prices.length-2);j++)

                           {

                             int a2=0;

                             int a4=0;

                           for(int k=i+1;k<j;k++)

                           {

                               if(prices[k]>a2)

                                   {

                                  a2= prices[k];

                               }

                           }

                           for(int m=j+1;m<=(prices.length-1);m++)

                               {

                              if(prices[m]>a4)

                                  {

                                 a4= prices[m];

                              }

                           }

                           if(max<(a2+a4-prices[i]-prices[j]))

                               {

                               max=a2+a4-prices[i]-prices[j];

                           }

                       }

                }

        }

        return max;

    }