Best Time to Buy and Sell Stock II -- LeetCode 原题链接:http://oj.leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/ 这道题跟Best Time to Buy and Sell Stock类似,求最大利润。区别是这里可以交易无限多次(当然我们知道交易不会超过n-1次,也就是每天都进行先卖然后买)。既然交易次数...
https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/ Design an algorithm to find the maximum profit. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times). Note: You may not engage in multiple transactions at t...
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). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again). ...
【leetcode】43-best-time-to-buy-and-sell-stock-iv 力扣 188. 买卖股票的最佳时机 IV 【leetcode】44-best-time-to-buy-and-sell-stock-with-cooldown 力扣 309. 买卖股票的最佳时机包含冷冻期 【leetcode】45-best-time-to-buy-and-sell-stock-with-cooldown 力扣 714. 买卖股票的最佳时机包含手续费 ...
问题https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock-ii/ 练习使用JavaScript解答 /** * @param {number[]} prices * @return {number} */ var maxProfit = function(prices) { if(prices.length < 2) return 0; ...
Leetcode: Best Time to Buy and Sell Stock II 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 ...
122. 买卖股票的最佳时机 II - 给你一个整数数组 prices ,其中 prices[i] 表示某支股票第 i 天的价格。 在每一天,你可以决定是否购买和/或出售股票。你在任何时候 最多 只能持有 一股 股票。你也可以先购买,然后在 同一天 出售。 返回 你能获得的 最大 利润 。 示
https://leetcode.com/problems/best-time-to-buy-and-sell-stock/解题思路1. 暴力 O(n^2) 复杂度,超时class Solution { public: int maxProfit(vector<int>& prices) { int maxVal = 0; int n = price…
122. 买卖股票的最佳时机 II - 给你一个整数数组 prices ,其中 prices[i] 表示某支股票第 i 天的价格。 在每一天,你可以决定是否购买和/或出售股票。你在任何时候 最多 只能持有 一股 股票。你也可以先购买,然后在 同一天 出售。 返回 你能获得的 最大 利润 。 示
My code: publicclassSolution{publicintmaxProfit(int[]prices){if(prices==null||prices.length<2)return0;boolean isInMode=false;inti=0;intmax=0;inttempMax=0;while(i<prices.length){for(intj=i+1;j<prices.length;j++){if(prices[j]<=prices[i]&&!isInMode){i=j;tempMax+=max;max=0;if(i...