122. Best Time to Buy and Sell Stock II
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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). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
题意:给定一个数组,数组下标代表第i天,数组值代表第i天的股票价格。
设计一个算法去找到最大的利润。可以多次买进多次卖出来完成交易。但是你不能在同一时间操作多次。(即在你再一次买这只股票之前你必须先卖掉)。
思路:
对于每一笔交易 假设是从i天买入 j天卖出 ,如果有j+1天的价格高于j天,所以肯定是从i到j+1利润更高,同样的如果有第i-1天价格比第i天更低,肯定是从i-1天到j天利润更高。分析可得,只要后项避前项大,大的这部分就一定可以变为利润。
代码:8ms
class Solution {public: int maxProfit(vector<int>& prices) { int maxProfit = 0; for(int i=1; i<prices.size(); i++){ if(prices[i]>prices[i-1]){ maxProfit += prices[i] - prices[i-1]; } } return maxProfit; }};
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