121. Best Time to Buy and Sell Stock
https://leetcode.com/problems/best-time-to-buy-and-sell-stock/#/description
Say you have an array for which the ith
element is the price of a given stock on day i.
If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.
Example 1:
Input: [7, 1, 5, 3, 6, 4]Output: 5max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)
Example 2:
Input: [7, 6, 4, 3, 1]Output: 0In this case, no transaction is done, i.e. max profit = 0.
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分析
- 初始化 buy_price = price[0]，作为买入价格
- 初始化 profit = 0
- 遍历price，若price[i]<buy_price, 则更新买入的价格 buy_price=price[i]
- 若 profit < price[i] - buy_price, 则
更新利润 profit = price[i] - buy_price
# Time O(n)
class Solution(object):
def maxProfit(self, prices):
"""
:type prices: List[int]
:rtype: int
"""
if len(prices) <= 1:
return 0
profit = 0
buy_price = prices[0]
for i in range(len(prices)):
if buy_price > prices[i]:
buy_price = prices[i]
if profit < prices[i] - buy_price:
profit = prices[i] - buy_price
return profit
122. Best Time to Buy and Sell Stock II
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 (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
分析
- 在 I 的基础上，允许多次买卖操作
- 只要有利润就可进行买卖操作
- 注意：不一定在第一天买 [2, 1, 4]-》3
- 所以先找到最小值作为买入
class Solution(object):
def maxProfit(self, prices):
"""
:type prices: List[int]
:rtype: int
"""
if len(prices) == 0:
return 0
profit = 0
buy_price = prices[0]
for i in range(1, len(prices)):
if prices[i] < buy_price:
buy_price = prices[i]
if prices[i] > buy_price:
profit += prices[i] - buy_price
buy_price = prices[i]
return profit
