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LeetCode | 0122. Best Time to Buy and Sell Stock II买卖股票的最佳时机 II【Python】

LeetCode 0122. Best Time to Buy and Sell Stock II买卖股票的最佳时机 II【Easy】【Python】【贪心】【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 as many transactions as you like (i.e., buy one and sell one share of the stock multiple times).

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: [7,1,5,3,6,4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
             Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 			  3.
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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.
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Example 3:

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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问题

力扣

给定一个数组,它的第 i 个元素是一支给定股票第 i 天的价格。

设计一个算法来计算你所能获取的最大利润。你可以尽可能地完成更多的交易(多次买卖一支股票)。

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: [7,1,5,3,6,4]
输出: 7
解释: 在第 2 天(股票价格 = 1)的时候买入,在第 3 天(股票价格 = 5)的时候卖出, 这笔交易所能获得利润 = 5-1 = 4 。
     随后,在第 4 天(股票价格 = 3)的时候买入,在第 5 天(股票价格 = 6)的时候卖出, 这笔交易所能获得利润 = 6-3 = 3 。
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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。
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示例 3:

输入: [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。
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思路

解法一

贪心

根据题意,是按照每天的顺序进行股票买入与卖出。
于是,只要后一天价格大于前一天价格,就可买入卖出这只股票。
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时间复杂度: O(len(prices)) 空间复杂度: O(1)

Python3代码
class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        # solution one: 贪心
        if not prices or len(prices) == 0:
            return 0
        profit = 0
        for i in range(len(prices)-1):
            if(prices[i+1] > prices[i]):
                profit += prices[i+1] - prices[i]
        return profit
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解法二

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 = +inf
因为 k 为正无穷,那么可以把 k 和 k-1 看成是一样的。
buy+sell = 一次完整的交易,这里把 sell 看成一次交易,所以第一行是 k-1。
dp[i][k][0] = max(dp[i-1][k][0], dp[i-1][k-1][1] + prices[i])
			= max(dp[i-1][k][0], dp[i-1][k][1] + prices[i])
dp[i][k][1] = max(dp[i-1][k][1], dp[i-1][k][0] - prices[i])

所以 k 对状态转移没有影响:
dp[i][0] = max(dp[i-1][0], dp[i-1][1] + prices[i])
dp[i][1] = max(dp[i-1][1], dp[i-1][0] - prices[i])

i = 0 时,dp[i-1] 不合法。
dp[0][0] = max(dp[-1][0], dp[-1][1] + prices[i])
         = max(0, -infinity + prices[i])
         = 0
dp[0][1] = max(dp[-1][1], dp[-1][0] - prices[i])
         = max(-infinity, 0 - prices[i]) 
         = -prices[i]
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空间复杂度: O(1)

Python3代码
class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        # solution two: 动态规划
        dp_i_0 = 0
        dp_i_1 = float('-inf')  # 负无穷
        for i in range(len(prices)):
            temp = dp_i_0
            # 昨天没有股票,昨天有股票今天卖出
            dp_i_0 = max(dp_i_0, dp_i_1 + prices[i])  # dp_i_0 和 dp_i_1 可以看成是变量,存储的都是上一次即昨天的值
            # 昨天有股票,昨天没有股票今天买入
            dp_i_1 = max(dp_i_1, temp - prices[i])
        return dp_i_0
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代码地址

GitHub链接

参考

一个方法团灭 6 道股票问题

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