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Question:
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).
Analysis：
问题描述：给出一个整数数组，其中第i个元素表示第i天得股价。
设计一个程序找到最大利润，最多只能进行两次交易。
思路一：暴力求解方法。一次循环分成两段，每段求最大利润，然后找到最大利润，时间复杂度为O（n2）.
思路二：Dynamic Programming的思想。首先正向遍历一遍，计算若是当前交易能够得到的profit；然后逆向遍历一遍（正向遍历与逆向遍历要求的东西是不一样的，正向是求当前如果进行交易的话能够得到的profit，而逆向遍历要求从i到最后能够获得的最大收益）。
Answer：
public class Solution { public static int maxProfit(int[] prices) { if(prices.length == 0 || prices.length == 1) return 0; int n = prices.length; //正向寻找最大利润 int low = prices[0]; int profit0 = 0; int [] pro = new int[n]; pro[0] = 0; for(int i=1; i<n; i++) { if(prices[i] < low) low = prices[i]; int temp = prices[i] - low; if(profit0 < temp) profit0 = temp; pro[i] = temp; } //逆向寻找最大利润 int high = prices[prices.length - 1]; int profit1 = 0; int[] pro1 = new int[n]; pro1[n-1] = 0; for(int i=n - 2; i>=0; i--) { if(high < prices[i]) high = prices[i]; int temp = high - prices[i]; if(profit1 < temp) profit1 = temp; pro1[i] = profit1; } int res = 0; for(int i=0; i<n; i++) { int temp = pro[i] + pro1[i]; //System.out.println("pro: "+pro[i]+" pro1: "+pro1[i] ); if(res < temp) res = temp; } return res; } }
LeetCode -- Best Time to Buy and Sell Stock III
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原文地址：http://www.cnblogs.com/little-YTMM/p/4821658.html
