标签：des style blog color strong os
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 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,布布扣,bubuko.com
【leetcode刷题笔记】Best Time to Buy and Sell Stock III
标签：des style blog color strong os
原文地址：http://www.cnblogs.com/sunshineatnoon/p/3856389.html
