121.Best Time to Buy and Sell Stock

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

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 <= 10⁵
0 <= prices[i] <= 10⁴

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;
}

Marsgoat 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;
    }
};

Yen-Chi ChenFri, 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