LeetCode算法题-Best Time to Buy and Sell Stock 这是悦乐书的第172次更新,第174篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第31题(顺位题号是121)。假设有一个数组,其中第i个元素是第i天给定股票的价格。如果只被允许完成最多一笔交易(即买入并卖出一股股票),请设计算法以找到最大利...
买卖股票系列 【leetcode】40-best-time-to-buy-and-sell-stock 力扣 121. 买卖股票的最佳时机 【leetcode】41-best-time-to-buy-and-sell-stock-ii 力扣 122. 买卖股票的最佳时机 II 【leetcode】42-best-time-to-buy-and-sell-stock-iii 力扣 123. 买卖股票的最佳时机 III 【leetcode】43-best-time-...
解题代码 (有缺陷): classSolution:defmaxProfit(self,prices:List[int])->int:## 双指针解法left,right=0,1# left=buy, right=sellmaxP=0whileright<len(prices):## 遍历整个 listifprices[right]>prices[left]:## 在存在赚钱机会的条件下profit=prices[right]-prices[left]maxP=max(maxP,profit)else:...
在计算第一个时,因为sell之前没有buy因此只能从第一个元素开始,buy的第一个设置为-prices[0],sell和presell设置为0,这样就可以计算第一个元素的buy。 另外还有一道比较类似的robber题目:https://leetcode.com/problems/house-robber/?tab=Description。思路还是应用一个序列,只不过stock里面要考虑以...
给定一个数组,它的第i个元素是一支给定股票第i天的价格。 如果你最多只允许完成一笔交易(即买入和卖出一支股票一次),设计一个算法来计算你所能获取的最大利润。 注意:你不能在买入股票前卖出股票。 示例1: 输入: [7,1,5,3,6,4]输出: 5解释: 在第 2 天(股票价格 = 1)的时候买入,在第 5 天(股票...
Can you solve this real interview question? Best Time to Buy and Sell Stock - You are given an array prices where prices[i] is the price of a given stock on the ith day. You want to maximize your profit by choosing a single day to buy one stock and choo
Leetcode: Best Time to Buy and Sell Stock 题目: 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 ...
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 0121. Best Time to Buy and Sell Stock买卖股票的最佳时机【Easy】【Python】【贪心】 Problem LeetCode Say you have an array for which theith element is the price of a given stock on dayi. If you were only permitted to complete at most one transaction (i.e., buy one and sell...
第一种思路:由于只能进行2次交易,那么就是选2个区间使得收益最大化,用dp[i][0]表示从0i天买入卖出一次股票得最大收益,dp[i][1]表示从in-1天买入卖出一次股票的最大收益。最后我们只需要枚举天数i,便可以获得最大收益,即dp[i][0]+dp[i+1][1]。由于还会出现只买入卖出一次的情况,那么最大收益为dp...