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
輸入會給我們一組整數陣列，從左到右分別是每天的股票價格。我們要做的便是讓程式決定怎麼樣買可以獲得最大的利潤（profit）。我們只能選擇一天買進、一天賣出。
如果無法獲得任何利潤，當然就是什麼也沒買，返回 0。
解題思路
解題的方法非常地單純：隨時紀錄當天價格，並比較是否比之前買入的價格低，若是有價格更低的天數，便隨時將買入的天數定為價格低的那天。
同時，每當遇到賣出價格高的日子，則隨時將賣出的利潤價格與歷史最高利潤價格做比較。
如此一來，等程式跑完後便能知道在哪天買入、哪天賣出的利潤最高。
這題題目是比較沒有實戰意味的。僅僅只是考好玩的感覺。
C++ 程式碼
class Solution { public: int maxProfit(vector<int>& prices) { // Init int profit=0; int buy=10001; // Determine for (int i=0; i<prices.size(); ++i) { if (prices[i] < buy) buy = prices[i]; if (profit < prices[i]-buy) profit = prices[i] - buy; } return profit; } };
我之所以將 buy 設定為 10001，是因為題目規定最大的股票價格是 10000，所以我才隨便選定了個更大的數字當作起始值。
Python 程式碼
class Solution: def maxProfit(self, prices: List[int]) -> int: # Init profit = 0 buy = 10001 # Determine for price in prices: if (price < buy): buy = price if (profit < price - buy): profit = price - buy return profit