image.png
0. 鏈接
1. 題目
Say you have an array prices 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.
Constraints:
1 <= prices.length <= 3 * 10 ^ 4
0 <= prices[i] <= 10 ^ 4
2. 思路1: 遍歷找波峰波谷
- 記錄buy_price, sell_price, max_profit
- 從左到右遍歷,
- 遇到price比buy_price小的, 則更新buy_price, - 遇到price比sell_price大的, 則更新sell_price
- 當(dāng)遇到price比上一個(gè)sell_price小, 說(shuō)明上一個(gè)sell_price就是波峰, 此時(shí)進(jìn)行一次結(jié)算 max_profit += sell_price - buy_price, 并重置buy_price=price, sell_price=None
- 時(shí)間復(fù)雜度: ```O(N)``
- 空間復(fù)雜度:
O(1)
3. 代碼
# coding:utf8
from typing import List
class Solution:
def maxProfit(self, prices: List[int]) -> int:
if len(prices) == 0:
return 0
buy_price = prices[0]
sell_price = None
max_profit = 0
for i in range(1, len(prices)):
price = prices[i]
if sell_price is None:
if price < buy_price:
buy_price = price
else:
sell_price = price
else:
if price < sell_price:
max_profit += sell_price - buy_price
buy_price = price
sell_price = None
else:
sell_price = price
if sell_price is not None:
max_profit += sell_price - buy_price
return max_profit
def my_test(solution, prices):
print('input: {}; output: {}'.format(prices, solution.maxProfit(prices)))
solution = Solution()
my_test(solution, [7, 1, 5, 3, 6, 4])
my_test(solution, [1, 2, 3, 4, 5])
my_test(solution, [7, 6, 4, 3, 1])
輸出結(jié)果
input: [7, 1, 5, 3, 6, 4]; output: 7
input: [1, 2, 3, 4, 5]; output: 4
input: [7, 6, 4, 3, 1]; output: 0
4. 結(jié)果
image.png
5. 思路2: 積分法
- 求所有呈上升趨勢(shì)的階段的增量和
- 時(shí)間復(fù)雜度
O(N) - 空間復(fù)雜度
O(1)
6. 代碼
# coding:utf8
from typing import List
class Solution:
def maxProfit(self, prices: List[int]) -> int:
max_profit = 0
for i in range(1, len(prices)):
if prices[i] > prices[i - 1]:
max_profit += prices[i] - prices[i - 1]
return max_profit
def my_test(solution, prices):
print('input: {}; output: {}'.format(prices, solution.maxProfit(prices)))
solution = Solution()
my_test(solution, [7, 1, 5, 3, 6, 4])
my_test(solution, [1, 2, 3, 4, 5])
my_test(solution, [7, 6, 4, 3, 1])
輸出結(jié)果為
input: [7, 1, 5, 3, 6, 4]; output: 7
input: [1, 2, 3, 4, 5]; output: 4
input: [7, 6, 4, 3, 1]; output: 0
7. 結(jié)果
image.png
