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:
和121. Best Time to Buy and Sell Stock类似,只不过可以买卖无数次。
思路:
既然可以买卖无数次,那么就可以买卖买卖……一直这样子循环下去,所以对于每次买卖,如果能够获得利润就交易,否则就不交易,利润为0.对于1,2,3这样的情况,也可以一直买卖,因为3-1=2-1+3-2.
代码:
class Solution {
public:
int maxProfit(vector<int>& prices) {
int n=prices.size();
int maxn=0;
for(int i=1;i<n;i++)
maxn+=max(prices[i]-prices[i-1],0);
return maxn;
}
};