LeetCode | 0123. Best Time to Bu

LeetCode 0123. Best Time to Buy and Sell Stock III买卖股票的最佳时机 III【Hard】【Python】【动态规划】

Problem

LeetCode

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

Design an algorithm to find the maximum profit. You may complete at most two transactions.

Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

Example 1:

Input: [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
             Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.

Example 2:

Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
             Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
             engaging multiple transactions at the same time. You must sell before buying again.

Example 3:

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

问题

力扣

给定一个数组，它的第 i 个元素是一支给定的股票在第 i 天的价格。

设计一个算法来计算你所能获取的最大利润。你最多可以完成 两笔 交易。

注意: 你不能同时参与多笔交易（你必须在再次购买前出售掉之前的股票）。

示例 1:

输入: [3,3,5,0,0,3,1,4]
输出: 6
解释: 在第 4 天（股票价格 = 0）的时候买入，在第 6 天（股票价格 = 3）的时候卖出，这笔交易所能获得利润 = 3-0 = 3 。
     随后，在第 7 天（股票价格 = 1）的时候买入，在第 8 天 （股票价格 = 4）的时候卖出，这笔交易所能获得利润 = 4-1 = 3 。

示例 2:

输入: [1,2,3,4,5]
输出: 4
解释: 在第 1 天（股票价格 = 1）的时候买入，在第 5 天 （股票价格 = 5）的时候卖出, 这笔交易所能获得利润 = 5-1 = 4 。   
     注意你不能在第 1 天和第 2 天接连购买股票，之后再将它们卖出。   
     因为这样属于同时参与了多笔交易，你必须在再次购买前出售掉之前的股票。

示例 3:

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

思路

动态规划

找到状态方程

dp[i][k][0] = max(dp[i-1][k][0], dp[i-1][k][1] + prices[i])
解释：昨天没有股票，昨天有股票今天卖出

dp[i][k][1] = max(dp[i-1][k][1], dp[i-1][k][0] - prices[i])
解释：昨天有股票，昨天没有股票今天买入

base case：
dp[-1][k][0] = dp[i][k][0] = 0
dp[-1][k][1] = dp[i][k][1] = -inf

k = 2
因为 k 为 2，所以要对 k 进行穷举。
dp[i][2][0] = max(dp[i-1][2][0], dp[i-1][2][1] + prices[i])
dp[i][2][1] = max(dp[i-1][2][1], dp[i-1][1][0] - prices[i])
dp[i][1][0] = max(dp[i-1][1][0], dp[i-1][1][1] + prices[i])
dp[i][1][1] = max(dp[i-1][1][1], -prices[i])

i = 0 时，dp[i-1] 不合法。
dp[0][1][0] = max(dp[-1][1][0], dp[-1][1][1] + prices[i])
            = max(0, -infinity + prices[i])
            = 0
dp[0][1][1] = max(dp[-1][1][1], dp[-1][1][0] - prices[i])
            = max(-infinity, 0 - prices[i]) 
            = -prices[i]
dp[0][2][0] = max(dp[-1][2][0], dp[-1][2][1] + prices[i])
            = max(0, -infinity + prices[i])
            = 0
dp[0][2][1] = max(dp[-1][2][1], dp[-1][2][0] - prices[i])
            = max(-infinity, 0 - prices[i])
            = -prices[i]

空间复杂度: O(1)

Python3代码

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        dp_i_2_0, dp_i_1_0 = 0, 0
        # dp_i_2_1, dp_i_1_1 = -prices[0], -prices[0]  # 会报错：list index out of range
        dp_i_2_1, dp_i_1_1 = float('-inf'), float('-inf')  # 负无穷
        for i in range(len(prices)):
            # 昨天没有股票，昨天有股票今天卖出
            dp_i_2_0 = max(dp_i_2_0, dp_i_2_1 + prices[i])
            # 昨天有股票，昨天没有股票今天买入
            dp_i_2_1 = max(dp_i_2_1, dp_i_1_0 - prices[i])
            # 昨天没有股票，昨天有股票今天卖出
            dp_i_1_0 = max(dp_i_1_0, dp_i_1_1 + prices[i])
            # 昨天有股票，昨天没有股票今天买入
            dp_i_1_1 = max(dp_i_1_1, -prices[i])
            
        return dp_i_2_0

代码地址

GitHub链接

参考

一个方法团灭 6 道股票问题