README_EN.md

121. Best Time to Buy and Sell Stock

中文文档

Description

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

Solutions

Python3

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        if not prices:
            return 0
        res = 0
        min_price = prices[0]
        for price in prices:
            min_price = min(min_price, price)
            res = max(res, price - min_price)
        return res

Java

class Solution {
    public int maxProfit(int[] prices) {
        if (prices == null) return 0;
        int res = 0;
        int min = Integer.MAX_VALUE;
        for (int price : prices) {
            min = Math.min(min, price);
            res = Math.max(res, price - min);
        }
        return res;
    }
}

JavaScript

/**
 * @param {number[]} prices
 * @return {number}
 */
const maxProfit = function (prices) {
  let min = prices[0];
  let profit = 0;
  for (let i = 0; i < prices.length; i++) {
    if (prices[i] < min) {
      min = prices[i];
    }
    if (profit < prices[i] - min) {
      profit = prices[i] - min;
    }
  }
  return profit;
};

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