[leetcode] 121. Best Time to Buy and Sell Stock @ python

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原题

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.

解法

edge case是当prices为空列表时, 返回0. 遍历价格列表, 如果遇到低价格, 则更新min_so_far, 如果遇到价格比min_so_far高, 则计算利润, 将可能的利润加到profits列表中, 最后求profits最大值.
Time: O(n)
Space: O(1)

代码

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        # edge case
        if not prices: return 0
        min_so_far = prices[0]
        profits = [0]
        for price in prices:
            # if we meet a low price, update the min_so_far
            if price < min_so_far:
                min_so_far = price
                continue
            if price > min_so_far:
                profits.append(price - min_so_far)
        return max(profits)