LeetCode 솔루션 분류
[2/25] 121. Best Time to Buy and Sell Stock
본문
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.
Constraints:
1 <= prices.length <= 105
0 <= prices[i] <= 104
Accepted
3.1M
Submissions
5.7M
Acceptance Rate
54.4%
태그
#Amazon, #Apple, #Adobe, #Facebook, #Microsoft, #Goldman Sachs, #Bolt, #Bloomberg, #Cisco, #Google, #Citadel, #Intel, #tcs, #PayTM
관련자료
-
링크
댓글 2
학부유학생님의 댓글
- 익명
- 작성일
class Solution:
def maxProfit(self, prices: List[int]) -> int:
min_price = float('inf')
max_profit = 0
for price in prices:
max_profit = max(max_profit, price - min_price)
min_price = min(min_price, price)
return max_profit
JJJJJJJJJJ님의 댓글
- 익명
- 작성일
class Solution:
def maxProfit(self, prices: List[int]) -> int:
profit = 0
buy = 10000
for x in prices:
if buy > x:
buy = x
profit = max(profit, x - buy)
return profit
def maxProfit(self, prices: List[int]) -> int:
profit = 0
buy = 10000
for x in prices:
if buy > x:
buy = x
profit = max(profit, x - buy)
return profit