121.Best Time to Buy and Sell Stock
===

###### tags: `Easy`,`Array`,`DP`

[121. Best Time to Buy and Sell Stock](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/)

### 題目描述

You are given an array `prices` where `prices[i]` is the price of a given stock on the i^th^ 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`.

### 範例

**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` <= 10^5^
* 0 <= `prices[i]` <= 10^4^

### 解答

#### Javascript

function maxProfit(prices) {
  let max = 0;
  let minPrice = prices[0];
  for (let i = 1; i < prices.length; i++) {
    if (prices[i] < minPrice) {
      minPrice = prices[i];
    }
    max = Math.max(max, prices[i] - minPrice);
  }

  return max;
}

>[name=Marsgoat][time= Dec 23, 2022]

#### C++

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int lower = INT_MAX, ans = 0;
        for (auto price : prices) {
            lower = min(lower, price);
            ans = max(ans, price - lower);
        }
        return ans;
    }
};

> [name=Yen-Chi Chen][time=Fri, Dec 23, 2022]

#### Python

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        lower, ans = math.inf, 0
        for price in prices:
            lower = min(lower, price)
            ans = max(ans, price - lower)
        return ans

> [name=Yen-Chi Chen][time=Fri, Dec 23, 2022]

### Reference

[回到題目列表](https://hackmd.io/@Marsgoat/leetcode_every_day)