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...
Can you solve this real interview question? Best Time to Buy and Sell Stock II - You are given an integer array prices where prices[i] is the price of a given stock on the ith day. On each day, you may decide to buy and/or sell the stock. You can only h
TagsCompanies 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...
188 Best Time to Buy and Sell Stock IV 买卖股票的最佳时机 IV 假设你有一个数组,其中第 i 个元素是第 i 天给定股票的价格。 设计一个算法来找到最大的利润。您最多可以完成 k 笔交易。 注意: 你不可以同时参与多笔交易(你必须在再次购买前出售掉之前的股票)。 详见:https://leetcode.com/problems/...
LeetCode Best Time to Buy and Sell Stock 买卖股票的最佳时机 (DP),题意:给定一个序列,第i个元素代表第i天这支股票的价格,问在最佳时机买入和卖出能赚多少钱?只买一次,且仅1股,假设本钱无限。思路:要找一个最低价的时候买入,在最高价的时候卖出利润会最大。
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
https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/ 因为可以进行无限次交易,并且在下一次buy之前必须已经sell。所以只需要把所有price曲线价格上涨的部分加起来就行。 class Solution(object): def maxProfit(self, prices): """
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...
LeetCode 0121. Best Time to Buy and Sell Stock买卖股票的最佳时机【Easy】【Python】【贪心】【动态规划】 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 ...
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:## 对于任意一个右指...