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Description of the problem
Given an array, which i-th element is a given stock i-day prices.
If you are only allowed up to complete a transaction (ie buying and selling a stock), to design an algorithm to compute the maximum profit you can get.
Note that you can not sell the stock before buying stocks.
Example 1:
输入: [7,1,5,3,6,4]
输出: 5
解释: 在第 2 天（股票价格 = 1）的时候买入，在第 5 天（股票价格 = 6）的时候卖出，最大利润 = 6-1 = 5 。
注意利润不能是 7-1 = 6, 因为卖出价格需要大于买入价格。
Example 2
输入: [7,6,4,3,1]
输出: 0
解释: 在这种情况下, 没有交易完成, 所以最大利润为 0。
Solution 1- Violence Act
Do two for loops
class Solution:
def maxProfit(self, prices: List[int]) -> int:
max_profit=0
for i in range(len(prices)):
for j in range(i):
max_profit=max(max_profit,prices[i]-prices[j])
return max_profit
Violence overtime
Once traversal solution 2-
After reading method is to write their own solution to a problem, some do not understand the original solution was added subsequent understanding, adding to sell the i-th day, then the maximum profit must be in [0, i-1] option to buy the lowest point in between, so through the array , sequentially obtains each difference selling point of time, selecting the maximum value
class Solution:
def maxProfit(self, prices: List[int]) -> int:
if prices==[]:
return 0
else:
min_price=prices[0]
max_profit=0
for i in range(len(prices)):
min_price=min(min_price,prices[i])
max_profit=max(max_profit,prices[i]-min_price)
return max_profit
