122. Best Time to Buy and Sell Stock II

最新推荐文章于 2024-01-03 12:56:44 发布

niunc

最新推荐文章于 2024-01-03 12:56:44 发布

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分类专栏： LeetCode

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博客围绕股票买卖问题展开，给定数组，其第i个元素为第i天股票价格。需设计算法找出最大利润，可进行多次买卖交易，但不能同时参与多笔交易，即卖后才能再买，还给出了示例。

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https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/description/

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.

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        Sum = 0
        if len(prices) < 2:
            return Sum
        ptr1 = 0
        ptr2 = 1
        while ptr2 < len(prices):
            if prices[ptr1] < prices[ptr2]:
                Sum += prices[ptr2] - prices[ptr1]
            ptr1 += 1
            ptr2 += 1
        return Sum