【leetcode刷题笔记】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).

---

题解：DP

从前往后扫描一遍数组，用LeftToRight[i]记录(0,i)得到的最大利润；

从后往前扫描一遍数组，用RightToLeft[i]记录(i,n-1)得到的最大利润；

最终的最大利润是max(LeftToRight[i]+RightToLeft[i])。

举个例子，如果prices = {1,2,3,2,5,7}，对应的

LeftToRight = {0,1,2,2,4,6}

RightToLeft = {6,5,5,5,2,0}

最终的最大利润就是2+5=7。表示在0~2(或0~3)这个区间取得利润2，并且在2~5(或者3~5)这个区间取得利润5，最终得到利润7.

代码如下：

 public class Solution {
     public int maxProfit(int[] prices) {
         if(prices == null || prices.length == 0)
             return 0;

         int[] LeftToRight = new int[prices.length];
         int[] RightToLeft = new int[prices.length];

         int minimal = prices[0];
         LeftToRight[0] = 0;
         for(int i = 1;i < prices.length;i++){
             minimal = Math.min(minimal, prices[i]);
             LeftToRight[i] = Math.max(LeftToRight[i-1], prices[i]-minimal);
         }

         int profit = 0;
         //From Right to left
         RightToLeft[prices.length-1] = 0;
         int maximal = prices[prices.length-1];
         for(int i = prices.length-2;i >= 0;i--){
             maximal = Math.max(prices[i], maximal);
             RightToLeft[i]= Math.max(RightToLeft[i+1], maximal-prices[i]);
             profit = Math.max(profit, LeftToRight[i] + RightToLeft[i] );
         }

         return Math.max(profit, LeftToRight[0]+RightToLeft[0]);
     }
 }