Question: Best Time to Buy and Sell Stock
Say you have an array for which the ith element is the price of a given stock on day i.
If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.
Analysis:
问题描述：给出一个数组，其中第i个元素表示第i天的股票售价。如果只允许一次买入卖出，请给出最大利润方案（即在第i天买入，第j天卖出使得利益最大）。
思路：要求保证在 i < j 的前提下，找出数组中最大和最小的元素，两者之间的差就是最大利润。因此用一个指针指向最低元素（若当前元素的值比史上最低值还低，则指向该元素），一个整数保存目前的利润（如果当前元素减去最低值还要大于已有利润，则更新一下）。
Answer：
public class Solution { public static int maxProfit(int[] prices) { if(prices.length == 0 || prices.length == 1) return 0; int low = prices[0]; int profile = 0; for(int i=1; i<prices.length; i++) { if(prices[i] < low) low = prices[i]; if(prices[i] - low > profile) profile = prices[i] - low; } return profile; } }
Question：Best Time to Buy and Sell Stock II
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 as many transactions as you like (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
Analysis:
题目描述：给出一个整数数组，第i个元素表示第i天得股票售价。 设计一个算法得到最大利润，与第一个版本不同的是，这次交易允许多次买卖，但是必须保证在买入股票之前要先卖出。
思路：开始想是否要用到动态规划的思想，后来发现，这个题目是要将原来的数组划分成一个个的小波峰波谷，而每次波峰与波谷之间的差值都可以作为利润。
Answer：
public class Solution { public int maxProfit(int[] prices) { if(prices.length == 0 || prices.length == 1) return 0; int profit = 0; for(int i=1; i<prices.length; i++) { int diff = prices[i] - prices[i-1]; if(diff > 0) { profit += diff; } } return profit; } }
Question：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).
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; } }