输入:[3,3,5,0,0,3,1,4]输出:6解释:在第 4 天(股票价格 = 0)的时候买入,在第 6 天(股票价格 = 3)的时候卖出,这笔交易所能获得利润 = 3-0 = 3 。 随后,在第 7 天(股票价格 = 1)的时候买入,在第 8 天 (股票价格 = 4)的时候卖出,这笔交易所能获得利润 = 4-1 = 3 。 示例2: 输...
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 (ie, you must sell the stock before you ...
classSolution {publicintmaxProfit(int[] prices) {if(prices ==null|| prices.length < 2)return0;//四个变量分别表示经过当前操作以后的profitintfirstBuy = Integer.MIN_VALUE, firstSell = 0;intsecondBuy = Integer.MIN_VALUE, secondSell = 0;for(intcurPrice : prices) { firstBuy= Math.max(first...
Note:You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again). Example 1: Input:prices = [3,3,5,0,0,3,1,4]Output:6Explanation:Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3. Then buy o...
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 ...
Problem List Problem List RegisterorSign in Premium Testcase Test Result Test Result Case 1Case 2Case 3 prices = [3,3,5,0,0,3,1,4] 9 1 2 3 › [3,3,5,0,0,3,1,4] [1,2,3,4,5] [7,6,4,3,1] Source
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)。
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 (ie, you must sell the stock before you buy again). classSolution{public:intmaxProfit(vector<int>&prices){// Start typing your C/C++...
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again). 分析: 依照题意,要求依据每天的股票价格。交易两次得到最大的利润。直接的想法就是利用分治法,从前往后循环依次将数组分为前后两个股票价格序列,分别得到最大的利润,将两者相加得到两次交易...
After you sell your stock, you cannot buy stock on next day. (ie, cooldown 1 day) Example: prices = [1, 2, 3, 0, 2] maxProfit = 3 transactions = [buy, sell, cooldown, buy, sell] 分析: 该问题使用动态规划方法解决,但是不容易获取动态规划的最优子结构。这里,我们借鉴于网站的方法,使...