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).
不要被ii,嚇到了,其實比前面那個還簡單。可以多次。那隻需把每次遞增的加起來即可!
// 累加所有遞增的值即可!!
int maxProfit(vector<int> &prices) {
if(prices.size() <= 1)return 0;
int profit_sum = 0;
for(int i = 1;i < prices.size();i++)
{
if(prices[i] - prices[i-1] > 0)
{
profit_sum += prices[i] - prices[i-1];
}
}
return profit_sum;
}