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 ...
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题目地址:https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock/ 解题思路一:暴力求解法 根据题目我们可以知道,我们知道最大利润“i”-“j”的差值就可以得到结果,所以时间复杂度为O(n*n),空间复杂度为O(1)。以下为源码: intmaxProfit(vector<int>&prices) {intsize=prices.size();if(size...
给定一个数组prices,它的第i个元素prices[i]表示一支给定股票第i天的价格。 你只能选择某一天买入这只股票,并选择在未来的某一个不同的日子卖出该股票。设计一个算法来计算你所能获取的最大利润。 返回你可以从这笔交易中获取的最大利润。如果你不能获取任何利润,返回0。
121. Best Time to Buy and Sell Stock 121 这个比较简单,大致思想: 1.如果sale-buy<=0,则该价格适合买入。 2.否则判断该价格卖出的价格是否最好。 代码如下: publicintmaxProfit(int[]prices){if(prices==null||prices.length==0){return0;}intres=0;intbuy=prices[0];for(inti=1;i<prices.length;i...
未经作者授权,禁止转载 LeetCode: Best To Buy And Sell Stock (I-IV). I will give best solutions I think for each question. 编程 科学 知识 科学科普 数学 计算 0评论 按热度排序 按时间排序 请先登录后发表评论 (・ω・)发表评论 表情 看看下面~来发评论吧lf...
class Solution { public: int maxProfit(vector<int>& prices) { if(prices.empty()){ return 0; } int minpri=prices[0]; int prof=0; for(int i=0;
You can only hold at most one share of the stock at any time. However, you can buy it then immediately sell it on the same day. Find and return the maximum profit you can achieve. 给定一个整数数组 prices,其中 prices[i] 表示第 i 天的股票价格。 每天,你可以选择买入和/或卖出股票。你...