LeetCode 122. 买卖股票的最佳时机 II
Table of Contents
中文版:
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 。
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: 在这种情况下, 没有交易完成, 所以最大利润为 0。
英文版:
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 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.
My answer:
Version 1 算法:快慢指针,i是慢指针,j是快指针。如果prices[i] > prices[j]则是降价(下坡),不适合卖出,此时快慢指针同时前进。否则涨价(上坡)适合卖出。如果 prices[j] < prices[j+1],说明继续上涨,j 后移动,i 不动,直到 j对应的值不再上涨。
class Solution: def maxProfit(self, prices: List[int]) -> int: # version 1: i = 0 j = 1 res = 0 while j < len(prices): if prices[i] > prices[j]: i += 1 j += 1 else: if j + 1 < len(prices) and prices[j] < prices[j+1]: j += 1 else: res += prices[j] - prices[i] j += 2 i = j - 1 return res
Version 2 算法:虽然题目中没提到,但我们是可以认为是可以在同一天又卖又买的,其实就相当于没操作(2块钱卖出再花2块钱买入)。所以并不需要找到价格的最高点和最低点再相减,而是只要在上升状态下就可以每天相减,如[1,2,3,4,5]的计算方法就是 (2-1)+(3-2)+(4-3)+(5-4)。[1,3,2,5,7] 的计算方法是(3-1)+(5-2)+(7=5) .参考下图理解:
# version 2: class Solution: def maxProfit(self, prices: List[int]) -> int: res = 0 for i in range(1, len(prices)): if prices[i] > prices[i-1]: res += prices[i] - prices[i-1] # else: # continue return res