出處： https://leetcode.com/explore/interview/card/top-interview-questions-easy/97/dynamic-programming/572/
You are given an array prices where prices[i] is the price of a given stock on the ith day.
You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock.
Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.
Example 1:
Input: prices = [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. Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.
Example 2:
Input: prices = [7,6,4,3,1] Output: 0 Explanation: In this case, no transactions are done and the max profit = 0.
Constraints:
1 <= prices.length <= 1050 <= prices[i] <= 104
var maxProfit = function(prices) {
let maxProfit = 0
let min = prices[0]
for(let i = 1; i < prices.length; i++) {
min = Math.min(prices[i], min)
maxProfit = Math.max(prices[i] - min, maxProfit)
}
return maxProfit
}其實這題想通之後並不是很難。因為他有先後次序的概念，所以必須大概的思維如下：
- 紀錄當前最小的金額，日後所有出現的金額 – 當前最小金額才有可機會出現最大利潤
- 取得當前最大利潤，用目前出現的金額 – 當前最小金額，如果利潤更高則覆寫最大利潤
這題用 js 怎麼寫效能都很差。日後研究出更快的寫法再來補充吧。
