- 分類：Array/DP
- 時(shí)間復(fù)雜度: O(n) 相當(dāng)于把所有節(jié)點(diǎn)都遍歷一遍
- 空間復(fù)雜度: O(n)->O(1)
121. Best Time to Buy and Sell Stock
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 (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.
Note that you cannot sell a stock before you buy one.
Example 1:
Input: [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.
Not 7-1 = 6, as selling price needs to be larger than buying price.
Example 2:
Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.
代碼：
代碼1DP：
class Solution:
def maxProfit(self, prices: List[int]) -> int:
if prices==None or len(prices)<2:
return 0
profits=[0 for i in range(len(prices))]
buyprice=prices[0]
for i in range(1,len(prices)):
if prices[i]<prices[i-1]:
buyprice=min(prices[i],buyprice)
profits[i]=max(prices[i]-buyprice,profits[i-1])
return profits[-1]
代碼2 O(1)：
class Solution:
def maxProfit(self, prices: List[int]) -> int:
if prices==None or len(prices)<2:
return 0
profit=0
buyprice=prices[0]
for i in range(1,len(prices)):
if prices[i]<prices[i-1]:
buyprice=min(prices[i],buyprice)
else:
profit=max(prices[i]-buyprice,profit)
return profit
代碼3 O(1)：
class Solution:
def maxProfit(self, prices: List[int]) -> int:
if prices==None or len(prices)<2:
return 0
profit=0
buyprice=prices[0]
for i in range(1,len(prices)):
if prices[i]<prices[i-1]:
buyprice=min(prices[i],buyprice)
else:
profit=max(prices[i]-buyprice,profit)
return profit
討論：
1.自己寫(xiě)了一個(gè)代碼怎么搞都通不過(guò)= 。=
2.最后看了GTH的思路，每次都覺(jué)得GTH思路好清晰好棒
3.這個(gè)題好像有點(diǎn)像linklist和tree的結(jié)合
4.每次都搞一條新的鏈條
