Problem

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 (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Example 1:

Input: [7, 1, 5, 3, 6, 4]
Output: 5

max. difference = 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

In this case, no transaction is done, i.e. max profit = 0.

即給定一個數組，按從前往後順序計算，求最大差。

Python 實現

'''
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 (ie, buy one and sell one share of the stock),
design an algorithm to find the maximum profit.

Example 1:
Input: [7, 1, 5, 3, 6, 4]
Output: 5

max. difference = 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

In this case, no transaction is done, i.e. max profit = 0.
'''
 


#author li.hzh

class Solution:
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        if prices is None or len(prices) <= 1:
            return 0
        min_val = prices[0]
        profit = 0
        for index in range(1, len(prices)):
 

            cur_val = prices[index]
            if cur_val < min_val:
                min_val = cur_val
            cur_profit = cur_val - min_val
            if cur_profit > profit:
                profit = cur_profit
        return profit


print(Solution().maxProfit([7, 1, 5, 3, 6, 4]))
print(Solution().maxProfit([7, 6, 4, 3, 1]))
print(Solution().maxProfit([7, 6, 7, 3, 5]))

分析

開始把問題想複雜了，寫了個遞迴的解法。程式碼很簡單，結果時間超時了。

考慮O(n)的解法，其實只要考慮最小值就好了。