編輯:關於Android編程
Best Time to Buy and Sell Stock II
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).
下面的程序一定要加上if(price.empty()) return 0;才能通過。int maxProfit(vector&prices) { if (prices.empty()) return 0; int profit = 0; for (int i = 0; i < prices.size()-1; i++) { if (prices[i] < prices[i+1]) profit += prices[i+1]-prices[i]; } return profit; }
也可以修改如下:
int maxProfit(vector好難發現的一個bug。size()默認返回的是無符號整數。&prices) { //if (prices.empty()) return 0; int profit = 0; for (int i = 0; i < int(prices.size())-1; i++) { if (prices[i] < prices[i+1]) profit += prices[i+1]-prices[i]; } return profit; }
int maxProfit2(vector&prices) { int profit = 0; for (int i = 1; i < prices.size(); i++) { if (prices[i-1] < prices[i]) profit += prices[i]-prices[i-1]; } return profit; }
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