0121. Best Time to Buy and Sell Stock.md

• Best Time to Buy and Sell Stock
• 难度：easy｜ 容易
• 相关知识点：Array Dynamic Programming greedy
• 题目链接：https://leetcode.com/problems/best-time-to-buy-and-sell-stock

题目描述：

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.

Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.

Example 1:

Input: prices = [7,1,5,3,6,4] Output: 5 Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5. Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell. Example 2:

Input: prices = [7,6,4,3,1] Output: 0 Explanation: In this case, no transactions are done and the max profit = 0.

Constraints:

1 <= prices.length <= 105 0 <= prices[i] <= 104

solution:

class Solution:
   def maxProfit(self, prices: List[int]) -> int:
       max_profit, min_price = 0, float('inf')
       for price in prices:
           min_price = min(min_price, price)
           profit = price - min_price
           max_profit = max(max_profit, profit)
       return max_profit

时间复杂度是O(n)，其中n是数组 prices 的长度

class Solution {
public:
    int maxProfit(std::vector<int>& prices) {
    if (prices.empty()) {
        return 0;
    };

    int min_price = INT_MAX; // Initialize the minimum price to a very large value
    int max_profit = 0; // Initialize the maximum profit to 0

    for (int price : prices) {
        // Update the minimum price if we find a lower price
        min_price = std::min(min_price, price);

        // Calculate the potential profit if we sell at the current price
        int potential_profit = price - min_price;

        // Update the maximum profit if the potential profit is greater
        max_profit = std::max(max_profit, potential_profit);
    };

    return max_profit;
};

};