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).
最直观的方法，用两层循环实现买时和卖时能赚的方法，这种方法在股票持续下跌的情况仍然遍历O(n^2),所以Time Limit Exceeded，
代码如下：（题目竟然理解错了，就是说手里不能同时拥有两支股票，还有要对输入参数进行判断）
int maxProfit(vector<int> &prices) { int sum = 0; for(int i=0;i<prices.size();i++) // i为买股票的时间 for(int j=i+1;j<prices.size();j++) //j为卖股票的时间 if(prices[j]>prices[i]) sum += (prices[j]-prices[i]); return sum; }
看看下面正确的代码：
int maxProfit(vector<int> &prices) { if(prices.size()<2) return 0; int sum = 0; for(int i=0;i<prices.size()-1;i++) if(prices[i]<prices[i+1]) sum += (prices[i+1]-prices[i]); return sum; }
[LeetCode] Best Time to Buy and Sell Stock II,布布扣,bubuko.com
[LeetCode] Best Time to Buy and Sell Stock II
原文：http://www.cnblogs.com/Xylophone/p/3792483.html