Leetcode 122 Best Time to Buy and Sell Stock Ⅱ

Time to Buy and Sell Stock Ⅱ"/>

Leetcode 122 Best Time to Buy and Sell Stock Ⅱ

Leetcode 122 Best Time to Buy and Sell Stock Ⅱ

题目描述

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.

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 areengaging 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.

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思路解析

• 动态规划：

与上一道题相比，这道题并不限制一个周期内的交易次数，列出状态转移方程：
d p [ i ] [ k ] [ 0 ] = m a x ( d p [ i − 1 ] [ k ] [ 0 ] , d p [ i − 1 ] [ k ] [ 1 ] + p r i c e s [ i ] ) d p [ i ] [ k ] [ 0 ] = m a x ( d p [ i − 1 ] [ k ] [ 1 ] , d p [ i − 1 ] [ k ] [ 0 ] − p r i c e s [ i ] ) dp[i][k][0] = max(dp[i-1][k][0], dp[i-1][k][1] + prices[i]) \\ dp[i][k][0] = max(dp[i-1][k][1], dp[i-1][k][0] - prices[i]) dp[i][k][0]=max(dp[i−1][k][0],dp[i−1][k][1]+prices[i])dp[i][k][0]=max(dp[i−1][k][1],dp[i−1][k][0]−prices[i])
代码如下：

class Solution:def maxProfit(self, prices: List[int]) -> int:length = len(prices)dp_i_k_0 = 0dp_i_k_1 = -1 * sys.maxsizefor i in range(length):temp = dp_i_k_0dp_i_k_0 = max(dp_i_k_0, dp_i_k_1 + prices[i])dp_i_k_1 = max(dp_i_k_1, temp - prices[i])return dp_i_k_0

时间复杂度为 O ( N ) O(N) O(N)