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题目描述:
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).
解题思路:相邻两元素,若下一个大于上一个,则profit加上这两个数的差,c++代码如下:
int maxProfit(vector<int>& prices) {
if(prices.size()<2)
return 0;
int profit = 0;
int temp;
for (int i =0;i<prices.size()-1;i++){
if (prices[i]<prices[i+1]){
profit+=prices[i+1] -prices[i];
}
}
return profit;
}
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