题目 Say you have an array for which the ith element is the price of a given stock on day i. Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times).…

题目 Say you have an array for which the ith element is the price of a given stock on day i. Design an algorithm to find the maximum profit. You may complete at most two transactions. Note: You may not engage in multiple transactions at the same time (…

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 the maximum profit. 分…

121题目描述: 解题:记录浏览过的天中最低的价格,并不断更新可能的最大收益,只允许买卖一次的动态规划思想. class Solution { public: int maxProfit(vector<int>& prices) { if(prices.size() == 0) return 0; int min_prices = prices[0]; int max_profit = 0; for(int i = 1 ; i < prices.size(); i++ ){ if…

Leetcode之动态规划(DP)专题-122. 买卖股票的最佳时机 II(Best Time to Buy and Sell Stock II) 股票问题: 121. 买卖股票的最佳时机 122. 买卖股票的最佳时机 II 123. 买卖股票的最佳时机 III 188. 买卖股票的最佳时机 IV 309. 最佳买卖股票时机含冷冻期 714. 买卖股票的最佳时机含手续费 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. 设计一个算法来计算你所能获取的最大利润.你可以尽可能地完成更…

转载请注明出处:z_zhaojun的博客 原文地址 题目地址 Best Time to Buy and Sell Stock II Say you have an array for which the ith element is the price of a given stock on day i. Design an algorithm to find the maximum profit. You may complete as many transactions as you lik…

先看一道leetcode题: Best Time to Buy and Sell Stock II Say you have an array for which the ith element is the price of a given stock on day i. Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one a…

122. Best Time to Buy and Sell Stock II Say you have an array for which the ith element is the price of a given stock on day i. Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell on…

121. Best Time to Buy and Sell Stock 题目的要求是只买卖一次,买的价格越低,卖的价格越高,肯定收益就越大 遍历整个数组,维护一个当前位置之前最低的买入价格,然后每次计算当前位置价格与之前最低价格的差值,获得最大差值即为结果 class Solution { public: int maxProfit(vector<int>& prices) { if(prices.empty()) ; ]; ; ;i < prices.size();i++){…

package y2019.Algorithm.array; /** * @ProjectName: cutter-point * @Package: y2019.Algorithm.array * @ClassName: MaxProfit2 * @Author: xiaof * @Description: 122. Best Time to Buy and Sell Stock II * Say you have an array for which the ith element is t…