(Leet Code c++)Best Time to Buy and Sell Stock(2)

122. Best Time to Buy and Sell Stock II

 

You are given an array prices where prices[i] is the price of a given stock on the ith day.

Find the maximum profit you can achieve. 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 simultaneously (i.e., you must sell the stock before you buy again).

 

Example 1:

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

 

Constraints:

• 1 <= prices.length <= 3 * 104
• 0 <= prices[i] <= 104

DP를 이용.

prices의 1번 인덱스부터 시작한다.

ans에 (i번째 값 - (i - 1)번째 값)과 0 중 큰 값을 더해나가면 된다.

 

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