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.

->找出最低價（買）和最高價（賣）

<解題>

1. 找出最低價（買）
2. 找出最高利潤Max(maxProfit, prices[i]-min)

class Solution {
    public int maxProfit(int[] prices) {
        int n = prices.length; 
        int min = Integer.MAX_VALUE;
        int maxProfit = 0;

        for (int i=0 ; i<n ;i++){
            if(prices[i]< min) min = prices[i];
            maxProfit = Math.max(maxProfit, prices[i]-min);
        }
        return maxProfit;
    }
}

Time Complexity: O(N) Space Complexity: O(1)

<補充>

Integer.MAX_VALUE //數字最大值