Leetcode 121. Best Time to Buy and Sell Stock

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.

Link: 121. Best Time to Buy and Sell Stock

Examples:

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.

思路: 股票在这个序列中只交易一次，也就是只买卖一次，求最大收益。两层循环逐个比较一定会超时，所以dp记录，记录什么呢？记录第i天后出现的最大值，这样就知道第i天能卖出的最高价格，最后一天之后的最大值是自己，依次向前推，倒数第二天后出现的最大值dp[n-2] = max(prices[n-2], dp[n-1])，直到第一天。

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        n = len(prices)
        if n <= 1: return 0
        res = 0
        dp = [0]*n # maximum value after i_th day
        dp[-1] = prices[-1]
        for i in range(2, n+1):
            dp[n-i] = max(prices[n-i], dp[n-i+1])
            res = max(res, dp[n-i]-prices[n-i])
        return res

另一种就是记录第i天的最好收益，注意不一定是第i天卖出，也可能是之前就卖出了，同时记录i天前的最低价格，这样dp[i] = max(dp[i-1], prices[i]-minPrices)

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        n = len(prices)
        if n <= 1: return 0
        minprice = prices[0]
        dp = [0]*n # maximum benefit at i_th day
        for i in range(1, n):
            minprice = min(minprice, prices[i])
            dp[i] = max(dp[i-1], prices[i]-minprice)
        return dp[-1]

日期: 2021-04-09