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...
LeetCode 122:买卖股票的最佳时机 II(Best Time to Buy and Sell Stock II) 交易次数限制:允许进行多次买卖操作。你可以尽可能地完成更多的交易,但是你不能同时参与多笔交易(即你必须在再次购买前出售掉之前的股票)。 目标:求得最大利润。这里的最大利润是所有可能交易的利润之和。 这个问题的核心在于利用每一次...
Say you have an array for which theith element is the price of a given stock on dayi. 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. 说白了,就是找一组数字里最大差值,并...
publicintmaxProfit(int[] prices){if(prices ==null|| prices.length <=1) {return0; }intmaxProfit = Integer.MIN_VALUE;intbuyPrice = prices[0];for(inti=1; i < prices.length; i++) {if(prices[i] > buyPrice) { maxProfit = Math.max(maxProfit, prices[i]-buyPrice); }else{ buyPrice ...
第一个:https://leetcode.com/problems/best-time-to-buy-and-sell-stock/?tab=Description 至多交易一次,返回最大利润。实质上是让求一个连续的子串,其最后一个元素和第一个元素只差最大。因为题目要求至多一次交易而不是必须一次,因此如果数组降序,可以选择不交易,所以理论上最小值是0,对应不...
https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/ 因为可以进行无限次交易,并且在下一次buy之前必须已经sell。所以只需要把所有price曲线价格上涨的部分加起来就行。 class Solution(object): def maxProfit(self, prices): """
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. 大致的意思就是输入一个股票的股价数组,第i个元素代表第i天的股价,只允许买入卖出一次,问最大的利润是多少?
188. Best Time to Buy and Sell Stock IV 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 mostktransactions. Note: You may not engage in multiple transactions at the same time (...
这道题目还有两个变体,Best Time to Buy and Sell Stock II和Best Time to Buy and Sell Stock III,虽然题目要求比较像,但是思路却变化比较大,Best Time to Buy and Sell Stock II可以交易无穷多次,思路还是比较不一样,而Best Time to Buy and Sell Stock III则限定这道题交易两次,其实还可以general到限定...
则当k > size / 2时,问题可以转化为:Best Time to Buy and Sell Stock II Python代码: classSolution:# @return an integer as the maximum profitdefmaxProfit(self, k, prices):size = len(prices)ifk > size /2:returnself.quickSolve(size, prices) ...