README_EN.md

122. Best Time to Buy and Sell Stock II

中文文档

Description

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

Solutions

Python3

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        if not prices:
            return 0
        res = 0
        for i in range(1, len(prices)):
            t = prices[i] - prices[i - 1]
            res += max(t, 0)
        return res

Java

class Solution {
    public int maxProfit(int[] prices) {
        if (prices == null) return 0;
        int res = 0;
        for (int i = 1, n = prices.length; i < n; ++i) {
            // 策略是所有上涨交易日都做买卖，所以下跌交易日都不做买卖
            int t = prices[i] - prices[i - 1];
            res += Math.max(t, 0);
        }
        return res;
    }
}

C++

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

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