9.11 练手腾讯50题 121

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本文探讨了Leetcode 121题“最佳买卖股票时机”的两种算法解决方案。首先介绍了一种递归方法，虽然实现了功能，但效率低下。随后提出了一种更优策略，通过一次遍历找到最大利润，大幅提升了算法性能。

Leetcode 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.

一开始用递归来写，想法是每次找剩余的最小值和另一边的最大值相减，代码如下：

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        n = len(prices)
        if(n == 0 or n ==1):
            return 0
        if(n == 2):
            if(prices[0] >= prices[1]):
                return 0
            else:
                return prices[1] - prices[0]
        buyVal = min(prices)
        buyIndex = prices.index(buyVal)
        if buyIndex == n-1:
            return self.maxProfit(prices[0:n-1])
        else:
            sellSec = prices[buyIndex+1:]
            sellVal = max(sellSec)
            output1 = sellVal - buyVal
            output2 = self.maxProfit(prices[0:buyIndex])
            return max(output1, output2)

时间beat 6% 空间 beat 5% ，可以说这个方法又慢又浪费空间了。。。

换了一个思路：遍历一次，每次查找当前区域的极小值和极大值，当碰到另一个极小值时更换极小值和覆盖区域。代码如下：

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        n = len(prices)
        if(n == 0 or n ==1):
            return 0
        maxProfit = 0
        minIndex = 0
        for i in range(1,n):
            if(prices[i] < prices[minIndex]):
                    minIndex = i
            else:
                maxProfit = max(maxProfit, prices[i]-prices[minIndex])
        return maxProfit

beat 99%+