Question

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 (i.e., buy one and sell one share of the stock multiple times).

Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

Example 1:

Input: [7,1,5,3,6,4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
             Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.

Example 2:

Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
             Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
             engaging multiple transactions at the same time. You must sell before buying again.

Example 3:

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

Answer

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int maxprofit = 0;
        if (prices.size() == 0) return maxprofit;
        for (int
 
 i = 1; i < prices.size(); i++) {
            if (prices[i-1] < prices[i]) {
                maxprofit += prices[i] - prices[i-1];
            } 
        }
        return maxprofit;
    }
    
};

TotalProfit=∑i​(height(peaki​)−height(valleyi​))

在上述情況下,如果我們跳過peak_i 和valley_j試圖獲得更多的利潤考慮分更多的高度不同,獲得的淨利潤將永遠比他們獲得的較小,因為C總是會比A + B較小。所以只需要見到區域性極小值就買入，之後遇到區域性極大值就賣出，以上演算法就採取該策略。