leetcode 121】的更多相关文章

121.买卖股票的最佳时机 题目 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. 如果你最多只允许完成一笔交易(即买入和卖出一支股票),设计一个算法来计算你所能获取的最大利润. 注意你不能在买入股票前卖出股票. 示例 1: 输入: [7,1,5,3,6,4] 输出: 5 解释: 在第 2 天(股票价格 = 1)的时候买入,在第 5 天(股票价格 = 6)的时候卖出,最大利润 = 6-1 = 5 . 注意利润不能是 7-1 = 6, 因为卖出价格需要大于买入价格. 示例 2: 输…
121. Best Time to Buy and Sell Stock 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 algor…
121. Best Time to Buy and Sell Stock 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 algor…
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. Exam…
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 (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit. No…
描述: 给一些列数字,表示每条股票的价格,如果可以买卖一次(不能同一天买和卖),求最大利益(即差最大). 其他三道问题是,如果能买卖无限次,买卖两次,买卖k次. 题一: 实质是求后面一个数减前一个数的最大差值. 维护一个最小值,和当前最大值.只需遍历一次,空间也是常数. int maxProfit(vector<int>& prices) { ) ; ]; ; ; i < prices.size(); i++) { ret = max(ret, prices[i] - min_)…
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++){…
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…
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. 解题思路…
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. Exam…