# Leetcode: 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).

```public int maxProfit(int[] prices) {
if (prices == null || prices.length <= 1) {
return 0;
}
int profit = 0;
for (int i = 0; i < prices.length; i++) {
if (i + 1 < prices.length && prices[i] < prices[i + 1]) {
profit += prices[i + 1] - prices[i];
}
}
return profit;
}```
Share

## 3 thoughts on “Leetcode: Best Time to Buy and Sell Stock II”

1. dude, you know tomorrow ‘s price, if it goes up, you buy today, right?

Isn’t this straight forward?

public int maxProfit(int[] prices) {
if(prices==null||prices.length==0)return 0;
int start=0;
for(int i=0;iprices[i]){
start+=prices[i+1]-prices[i];
}
}
return start;
}

2. my code is strip off after I post. but you know the idea, if prices[i+1] greater than prices[i]. just plus the diff (prices[i+1]-prices[i])