123. Best Time to Buy and Sell Stock III 题目链接:https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/description/ 题目标签:Array,DP 与 122. Best Time to Buy and Sell Stock II 相比,此题可以最多进行两次交易。所以我们在
同样的道理来求f2[i], 只是这里求得是当前max减去当前价格的最大值, f2[i]与f2[i+1]有关 class Solution: # @param prices, a list of integer # @return an integer def maxProfit(self, prices): length=len(prices) if length==0: return 0 f1=[0 for i in range(length)] f2=[0 for i ...
Can you solve this real interview question? Best Time to Buy and Sell Stock III - You are given an array prices where prices[i] is the price of a given stock on the ith day. Find the maximum profit you can achieve. You may complete at most two transacti
Leetcode - 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 transactio......
这道题是Best Time to Buy and Sell Stock的扩展,现在我们最多可以进行两次交易。我们仍然使用动态规划来完成,事实上可以解决非常通用的情况,也就是最多进行k次交易的情况。 这里我们先解释最多可以进行k次交易的算法,然后最多进行两次我们只需要把k取成2即可。我们还是使用“局部最优和全局最优解法”。我们维护...
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again). 题意及分析:给出一个数组,代表一支股票在某天的价格,求能得到的最大利润。最多可以进行两次交易,同时不能同时持有两支股票。
[LeetCode]Best Time to Buy and Sell Stock III 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 (i.e., you must sell the stock before you buy again). Example 1: Input: [3,3,5,0,0,3,1,4] Output: 6 Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3. Then ...
Best Time to Buy and Sell Stock 相对比较简单的方法是用DP,思路是对于每个i都求出从0到i区间内的最大获益,而对于i+1只需要比较第i+1天的价格和前i天最低价的关系,就可以直接求出0到i+1天区间内的最大获益。也就是说对0到i天的最大获益的计算复杂度是O(1),总体复杂度是O(n)。
这道是买股票的最佳时间系列问题中最难最复杂的一道,前面两道Best Time to Buy and Sell Stock 买卖股票的最佳时间和Best Time to Buy and Sell Stock II 买股票的最佳时间之二的思路都非常的简洁明了,算法也很简单。而这道是要求最多交易两次,找到最大利润,还是需要用动态规划Dynamic Programming来解,而这里我...