[leetcode]121. Best Time to Buy and Sell Stock[facebook]
2017-12-11 11:09
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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:
Example 2:
我的做法:
我以为是用动态规划,但实际上不是。。。
是一道要用数学思考的题目。
而且get到的一点就是,可以不用是用if a < b: a=b 这种操作,直接a = min(a,b) ,max同理。
目前的做法是,记住遇到的最小值,每次更新profit,然后比较profit和max profit。
缺点:
慢
改进1:
如果用if a < b: a=b 这种操作,能快一点。
改进2:
上面是拿max profit 和min price和循环的元素做比较,改进是拿循环的元素和min price做比较。
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.
我的做法:
我以为是用动态规划,但实际上不是。。。
是一道要用数学思考的题目。
而且get到的一点就是,可以不用是用if a < b: a=b 这种操作,直接a = min(a,b) ,max同理。
目前的做法是,记住遇到的最小值,每次更新profit,然后比较profit和max profit。
缺点:
慢
改进1:
如果用if a < b: a=b 这种操作,能快一点。
改进2:
上面是拿max profit 和min price和循环的元素做比较,改进是拿循环的元素和min price做比较。
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