(Leet Code c++)Best Time to Buy and Sell Stock

121. Best Time to Buy and Sell Stock

 

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 <= 105
• 0 <= prices[i] <= 104

 

주어진 prices 배열에서 가장 작은 값 min을 갱신해 나가야 한다.

처음 시작할 때에는 prices[0]의 값을 min으로 잡고 시작한다.

1 ~ prices.size()까지 for문을 반복.

만약 i번째의 수가 min보다 작으면 min 값을 prices[i]로 갱신해준다.

그렇지 않고 i번째의 수가 min보다 크면 prices[i] - min을 ans와 비교하여 더 큰 값을 ans에 넣어주면 된다.

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int ans = 0;
        int min = prices[0];
        
        for(int i = 1; i < prices.size(); i++){
            if(prices[i] < min){
                min = prices[i];
            }
            else{
                ans = ans < prices[i] - min ? prices[i] - min : ans;
            }
        }
        
        return ans;
    }
};