-
Notifications
You must be signed in to change notification settings - Fork 0
/
121. Best Time to Buy and Sell Stock.py
51 lines (40 loc) · 1.79 KB
/
121. Best Time to Buy and Sell Stock.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
"""
https://leetcode.com/problems/best-time-to-buy-and-sell-stock/submissions/
You are given an array prices where prices[i] is the price of a given stock on the ith day.
You want to maximize your profit by choosing a single day to buy one stock and choosing a different day
in the future to sell that stock.
Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.
Example 1:
Input: prices = [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.
Example 2:
Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transactions are done and the max profit = 0.
"""
class Solution(object):
def maxProfit(self, prices):
# это решение знает о максимальной цене в будущем
mx = prices[-1]
right_max = [mx] # занимает память
for i in range(len(prices) - 1 - 1, 0, -1):
mx = max(mx, prices[i])
right_max.insert(0, mx)
profit = 0
for i in range(len(prices) - 1):
profit = max(profit, right_max[i] - prices[i])
return profit
# интересное
# def maxProfit(self, prices):
# а это решение знает о минимальном цене в прошлом
# max_profit = 0
# min_price = float('inf')
# for price in prices:
# max_profit = max(max_profit, price - min_price)
# min_price = min(min_price, price)
# return max_profit
solution = Solution()
assert solution.maxProfit([7, 1, 5, 3, 6, 4]) == 5
assert solution.maxProfit([7, 6, 4, 3, 1]) == 0