0121-买卖股票最佳时机(0121 – the best time to buy and sell stocks)

给定一个数组 prices ,它的第 i 个元素 prices[i] 表示一支给定股票第 i 天的价格。

你只能选择 某一天 买入这只股票,并选择在 未来的某一个不同的日子 卖出该股票。设计一个算法来计算你所能获取的最大利润。

返回你可以从这笔交易中获取的最大利润。如果你不能获取任何利润,返回 0 。

示例 1:

输入:[7,1,5,3,6,4]
输出:5
解释:在第 2 天(股票价格 = 1)的时候买入,在第 5 天(股票价格 = 6)的时候卖出,最大利润 = 6-1 = 5 。
注意利润不能是 7-1 = 6, 因为卖出价格需要大于买入价格;同时,你不能在买入前卖出股票。
示例 2:

输入:prices = [7,6,4,3,1]
输出:0
解释:在这种情况下, 没有交易完成, 所以最大利润为 0。

提示:

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

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock

参考:

  • https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock/solution/dai-ma-sui-xiang-lu-121-mai-mai-gu-piao-odhle/

python

# 0121.买卖股票最佳时机

# 贪心
class Solution:
    def maxProfit(self, prices:[int]) -> int:
        low = float("INF")
        res = 0
        for i in range(len(prices)):
            low = min(low, prices[i])
            res = max(res, prices[i]-low)
        return res

golang

package greedy

import (
	"fmt"
	"math"
)


// 贪心算法
func maxProfit(prices []int) int {
	low := math.MaxInt64
	res := 0
	for _,v := range prices {
		low = min(low, v)
		res = max(res, v-low)
	}
	return res
}

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

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

————————

Given an array price, its second   I elements   Prices [i] represents the price of a given stock on day I.

You can only choose to buy this stock one day and sell it on a different day in the future. Design an algorithm to calculate the maximum profit you can make.

Return the maximum profit you can make from this transaction. If you can’t make any profit, return 0.

Example 1:

Input: [7,1,5,3,6,4]
Output: 5
Explanation: buy on day 2 (stock price = 1) and sell on day 5 (stock price = 6). Maximum profit = 6-1 = 5.
Note that the profit cannot be 7-1 = 6, because the selling price needs to be greater than the buying price; At the same time, you can’t sell stocks before buying.
Example 2:

Input: prices = [7,6,4,3,1]
Output: 0
Explanation: in this case, no transaction is completed, so the maximum profit is 0.

Tips:

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

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock

reference resources:

  • https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock/solution/dai-ma-sui-xiang-lu-121-mai-mai-gu-piao-odhle/

python

# 0121.买卖股票最佳时机

# 贪心
class Solution:
    def maxProfit(self, prices:[int]) -> int:
        low = float("INF")
        res = 0
        for i in range(len(prices)):
            low = min(low, prices[i])
            res = max(res, prices[i]-low)
        return res

golang

package greedy

import (
	"fmt"
	"math"
)


// 贪心算法
func maxProfit(prices []int) int {
	low := math.MaxInt64
	res := 0
	for _,v := range prices {
		low = min(low, v)
		res = max(res, v-low)
	}
	return res
}

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

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