【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).

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 left
18         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 }

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