#leetcode# Best Time to Buy and Sell Stock II

題目：

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).

思路：

先得到b[i] = a[i] - a[i - 1]的這個數組, 然後問題就可以轉換爲每天要不要取 b[i] 這個數字了~~~

代碼:

int maxProfit(vector<int>& prices) {
        int n = prices.size();
        int ret = 0;
        for(int i = 1;i < n;i ++) {
            int add = prices[i] - prices[i - 1];
            if(add > 0) {
                ret += add;
            }
        }
       return ret;
    }