题目一
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.
答案一
def maxProfit(self, prices: List[int]) -> int:
if len(prices)<2:
return 0
# 暴力方法 O(N^2)
# profit = 0
# b_min = prices[0]
# for b in range(len(prices)-1):
# if (prices[b] < b_min) or (b==0):
# b_min = prices[b]
# s = max(prices[b+1:])
# if s - prices[b] > profit:
# profit = s - prices[b]
# else:
# pass
# return profit
# 非暴力方法 O(N)
mini_buy = 2**31
profit = 0
for i in range(0,len(prices)):
if prices[i] < mini_buy:
mini_buy = prices[i]
elif prices[i] - mini_buy > profit:
profit = prices[i] - mini_buy
return profit题目二
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 (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.
答案二
这个题需要下面的数学知识。
对于一个波峰波谷图来说，只要出现波峰就卖，所得的利润将比跨波峰出售大。即 C < A+B
其次，还可以将图中的折线平移，可以看出，只要后一日比前一日的价格高，我们都可以认为是在前一日买入，在后一日卖出。（同一日买入卖出，可以被视作这一日无操作
因此答案为：
class Solution:
def maxProfit(self, prices: List[int]) -> int:
if len(prices)<2:
return 0
profit = 0
for i in range(len(prices)-1):
if prices[i+1]>prices[i]:
profit += (prices[i+1]-prices[i])
return profit