Leetcode 121 Best Time to Buy and Sell Stock

Time to Buy and Sell Stock"/>

Leetcode 121 Best Time to Buy and Sell Stock

Leetcode 121 Best Time to Buy and Sell Stock

题目描述

Say you have an array for which the ith element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Note that you cannot sell a stock before you buy one.

Example 1:

Input: [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.Not 7-1 = 6, as selling price needs to be larger than buying price.

Example 2:

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

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思路解析

• 思路一：暴力法

题目要求得到买卖股票的最大差值，用暴力法暴力循环可以得到。代码如下：

class Solution:def maxProfit(self, prices: List[int]) -> int:max = 0length = len(prices)for i in range(length - 1):for j in range(i + 1, length):if prices[j] - prices[i] > max:max = prices[j] - prices[i]            return max

时间复杂度为 O ( N 2 ) O(N^2) O(N2)，并不能通过，但是我们可以对上述代码做一定的改进，剪去一些不必要的分支，提高效率，代码如下：

class Solution:def maxProfit(self, prices: List[int]) -> int:max = 0length = len(prices)for i in range(length - 1):if prices[i] >= prices[i + 1]:continuefor j in range(i + 1, length):if prices[j] <= prices[j - 1]:continueif prices[j] - prices[i] > max:max = prices[j] - prices[i]            return max

• 思路二：动态规划

class Solution:def maxProfit(self, prices: List[int]) -> int:min_prices = sys.maxsizemax_profit = 0length = len(prices)for i in range(length):if prices[i] <= min_prices:min_prices = prices[i]elif prices[i] - min_prices > max_profit:max_profit = prices[i] - min_pricesreturn max_profit

时间复杂度为 O ( N ) O(N) O(N)

动态规划这一块，还是不太行。

参考