You are given an array prices where prices[i] is the price of a given stock on the ith day.
You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock.
\n\nReturn the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.
\n
Example 1:
\n\n\nInput: prices = [7,1,5,3,6,4]\nOutput: 5\nExplanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.\nNote that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.\n\n\n
Example 2:
\n\n\nInput: prices = [7,6,4,3,1]\nOutput: 0\nExplanation: In this case, no transactions are done and the max profit = 0.\n\n\n
\n
Constraints:
\n\n1 <= prices.length <= 1050 <= prices[i] <= 104给定一个数组 prices ，它的第 i 个元素 prices[i] 表示一支给定股票第 i 天的价格。
你只能选择 某一天 买入这只股票，并选择在 未来的某一个不同的日子 卖出该股票。设计一个算法来计算你所能获取的最大利润。
\n\n返回你可以从这笔交易中获取的最大利润。如果你不能获取任何利润，返回 0 。
\n\n
示例 1：
\n\n\n输入：[7,1,5,3,6,4]\n输出：5\n解释：在第 2 天（股票价格 = 1）的时候买入，在第 5 天（股票价格 = 6）的时候卖出，最大利润 = 6-1 = 5 。\n 注意利润不能是 7-1 = 6, 因为卖出价格需要大于买入价格；同时，你不能在买入前卖出股票。\n\n\n
示例 2：
\n\n\n输入：prices = [7,6,4,3,1]\n输出：0\n解释：在这种情况下, 没有交易完成, 所以最大利润为 0。\n\n\n
\n\n
提示：
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