121. Best Time to Buy and Sell Stock

題目：
給一個股價序列，求一次買入和賣出可以得到的最大利潤

範例：

• 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.

想法：
最大利潤 = 最便宜的買入點 + 最高的賣出點（在買入日之後）
邊走陣列邊更新數值

程式碼：

class Solution {
    public int maxProfit(int[] prices) {
        int minPrice = Integer.MAX_VALUE; //初始為最大值
        int maxProfit = 0;                //初始利潤0

        for (int price : prices) {        //走訪每天股價
            if (price < minPrice) {       //若遇到更低股價 → 更新買入價
                minPrice = price;
            } else if (price - minPrice > maxProfit) { //計算今天賣的利潤
                maxProfit = price - minPrice;          //若利潤更高 → 更新最大利潤
            }
        }
        return maxProfit;
    }
}

實際操作：

prices = [7, 1, 5, 3, 6, 4]

STEP1:
minPrice = 7
profit = 7-7=0
maxProfit = 0

STEP2:
minPrice = 1 (遇到更低價)
profit = 1-1=0
maxProfit = 0

STEP3:
minPrice = 1
profit = 5-1=4
maxProfit = 4 (更新)

STEP4:
minPrice = 1
profit = 3-1=2
maxProfit = 4 (不變)

STEP5:
minPrice = 1
profit = 6-1=5
maxProfit = 5 (更新)

STEP6:
minPrice = 1
profit = 4-1=3
maxProfit = 5 (不變)

最終 maxProfit = 5