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-...
Say you have an array for which theith element is the price of a given stock on dayi. Design an algorithm to find the maximum profit. You may complete at mosttwotransactions. Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before yo...
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): """ :type prices: List[int] :rtype: int """ maxprofit = 0...
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 ...
You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again). The transaction fee is only charged once for each stock purchase and sale. Example 1: Input:prices = [1,3,2,8,4,9], fee = 2Output:8Explanation:The maximum profit can be...
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】45-best-time-to-buy-and-sell-stock-with-cooldown 力扣 714. 买卖股票的最佳时机包含手续费 开源地址 为了便于大家学习,所有实现均已开源。欢迎 fork + star~ https://github.com/houbb/leetcode 122. 买卖股票的最佳时机 II 给你一个整数数组 prices ,其中 prices[i] 表示某支股票第 i 天...
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. code : 朴素的算法,对每一个i ,计算 最大 的 prices[j] (j>i) 来维护最大的差值,但是这样的复杂度是O(n^2),会TLE ...
遍历价格vector,把每段上升期间的利润加起来,注意最后结束时记得再计算最后一次的利润 class Solution { // 把每段上升期间的利润加起来 // minVal: 低谷值 // maxVal: 峰值 // profit: 利润 public: int maxProfit(vector