题目信息 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 choosing a different day in the future to sell that stock. Return the maximum profit you can achieve ...
package day_40;// 个人感觉这个题目和 之前的Stock2是一样的// 上面那句是错的,实际上两个题目是不同的思路。publicclassBuySellStock3{publicintmaxProfit(int[]prices){int[]local=newint[3];// k+1int[]global=newint[3];for(inti=1;i<prices.length;i++){intdiff=prices[i]-prices[i-1];for...
第一种情况:就代表当前持有股票。用f[i][0]来记录,而f[i]就代表在第i天获得的最大收益是多少。 第二种情况:第i天买了股票,处于冷冻期(无法买入),用f[i][1] 来记录 第三种情况:没有持有股2票,也没有冷冻期(就是可以买入的状态)。用f[i][2]来表示。 我觉得作者这种思路跟我一开始想的没太大区...
right = 0, 1 # left=buy, right=sell maxP = 0 while right < len(prices): ## 遍历整个 list if prices[right] > prices[left]: ## 在存在赚钱机会的条件下 profit = prices[right] - prices[left] maxP = max(maxP, profit) else: ## 对于任意一个右指针日期,我们都要回顾它后面的所有...
309. Best Time to Buy and Sell Stock with Cooldown 最佳买卖股票时机含冷冻期 Title 给定一个整数数组,其中第 i 个元素代表了第 i 天的股票价格 。 设计一个算法计算出最大利润。在满足以下约束条件下,你可以尽可能地完成更多的交易(多次买卖一支股票):...
5 things to know when top execs buy or sell stock
best time to buy and sell stock iii: 最多允许买卖两次 对于i: 思路就是在第i个位置抛出时,希望在0~i-1个位置上的最低价买入,才能使得第i个位置的收益最大。然后比较所有可以抛出的位置i,返回最大的收益。 1 2 3 4 5 6 7 8 intmaxP = 0, minV = prices[0]; ...
若prices[sell] < prices[buy],则将buy指向sell当前位置(buy = sell),否则计算当前股票买卖收益,并和之前计算的收益比较,取最大的值。 1//双指针法求股票买卖最佳时机问题23intmaxProfit_doublePtr(vector<int>&prices)4{5intbuy=0,sell=0;6intmaxProfit =0;7for(;sell<prices.size();++sell)8{9if(pri...
AMC Stock Now: Is It A Buy Now? Or A Sell? AMC still sits more than 98% below its split-adjusted high of 393.65 set on June 2, 2021. However, the stock still needs to show more convincing signs of bottoming out. One could draw a trendline from the an early May peak of 5.96 on...
You can buy or sell stock on your own by opening a brokerage account with one of the many brokerage firms. After opening your account, connect it with your bank checking account to make deposits, which are then available for you to invest in. ...