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:
        res, mi = 0, prices[0]
        for price in prices[1:]:
            res = max(res, price - mi)
            mi = min(mi, price)
        return res

Java

class Solution {
    public int maxProfit(int[] prices) {
        int res = 0, mi = prices[0];
        for (int i = 1; i < prices.length; ++i) {
            res = Math.max(res, prices[i] - mi);
            mi = Math.min(mi, prices[i]);
        }
        return res;
    }
}

C++

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int res = 0, mi = prices[0];
        for (int i = 1; i < prices.size(); ++i) {
            res = max(res, prices[i] - mi);
            mi = min(mi, prices[i]);
        }
        return res;
    }
};

Go

func maxProfit(prices []int) int {
	res, mi := 0, prices[0]
	for i := 1; i < len(prices); i++ {
		res = max(res, prices[i]-mi)
		mi = min(min, prices[i])
	}
	return res
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

JavaScript

/**
 * @param {number[]} prices
 * @return {number}
 */
var maxProfit = function(prices) {
    let res = 0;
    let mi = prices[0];
    for (let i = 1; i < prices.length; ++i) {
        res = Math.max(res, prices[i] - mi);
        mi = Math.min(mi, prices[i]);
    }
    return res;
};

C#

public class Solution {
    public int MaxProfit(int[] prices) {
        int res = 0, mi = prices[0];
        for (int i = 1; i < prices.Length; ++i)
        {
            res = Math.Max(res, prices[i] - mi);
            mi = Math.Min(mi, prices[i]);
        }
        return res;
    }
}

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