前言

大家好，我是毛毛。ヾ(´∀ ˋ)ﾉ
沒錯~ 今天跟昨天是同一題(絕對不是想偷懶XD
想嘗試看看改C++寫
那就開始今天的解題吧~

Question

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 <= 10^5
• 0 <= prices[i] <= 10^4

Think

給一個陣列prices，其中每個數字代表的是美一天的股票價格。目標當然就是買低賣高，找出其中價差最高的~

• 用兩個變數buy跟sell當兩個index~一開始buy是0；sell是1，只有在sell那一天的價格大於buy那天的價格才去計算&跟max_profit比較誰比較大。

• 小於等於的話，就讓buy從sell現在的位置在繼續做以上的動作~

• 一樣XD

Code

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        
        // for (int i=0 ; i<prices.size() ; i++){
        //     cout << prices[i] << endl;
        // }
        
        int buy = 0;
        int sell = 1;
        int max_profit = 0;
        
        while (sell < prices.size()){
            if (prices[buy] < prices[sell]){
                max_profit = max((prices[sell]-prices[buy]), max_profit);
            } else {
                buy = sell;
            }
            sell++;
        }
        
        
        return max_profit;
    }
}; 

Submission

今天就到這邊啦~
大家明天見