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【leetcode刷题笔记】Best Time to Buy and Sell Stock III

2024-07-13 02:15:36   219人阅读

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：

 1 public class Solution { 2     public int maxProfit(int[] prices) { 3         if(prices == null || prices.length == 0) 4             return 0; 5          6         int[] LeftToRight = new int[prices.length]; 7         int[] RightToLeft = new int[prices.length]; 8          9         int minimal = prices[0];10         LeftToRight[0] = 0;11         for(int i = 1;i < prices.length;i++){12             minimal = Math.min(minimal, prices[i]);13             LeftToRight[i] = Math.max(LeftToRight[i-1], prices[i]-minimal);14         }15         16         int profit = 0;17         //From Right to left18         RightToLeft[prices.length-1] = 0;19         int maximal = prices[prices.length-1];20         for(int i = prices.length-2;i >= 0;i--){21             maximal = Math.max(prices[i], maximal);22             RightToLeft[i]= Math.max(RightToLeft[i+1], maximal-prices[i]);23             profit = Math.max(profit, LeftToRight[i] + RightToLeft[i] );24         }25         26         return Math.max(profit, LeftToRight[0]+RightToLeft[0]);27     }28 }


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