Say you have an array for which the ith element is the price of a given stock on day i.
If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.
Example 1:
Input: [7, 1, 5, 3, 6, 4]
Output: 5
max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)
Example 2:
Input: [7, 6, 4, 3, 1
]Output: 0
In this case, no transaction is done, i.e. max profit = 0.
給定一個數組，第i個元素為股票在第i天的價格。如果僅允許一次交易，設計算法找出最大的利潤。

算法分析

方法一：

嵌套for循環，找出最大的利潤。但是時間復雜度為O（n^2）

for(int i = 0; i < prices.length - 1; i ++)
    for(int j = i + 1; j < prices.length; j ++)

方法二：

一個for循環做兩次判斷。先判斷是不是最小的價格，如果不是最小的價格判斷是不是最大的利潤

Java代碼

public class Solution {
    public int maxProfit(int[] prices) {
        int minPrice = Integer.MAX_VALUE;
        int maxProfit = 0;
        for (int i = 0; i < prices.length; i++) {
            if (prices[i] < minPrice) {//找出最小的price
                minPrice = prices[i];
            } else if (prices[i] - minPrice > maxProfit) {//如果不是最小的price，是否是最大的利潤
                maxProfit = prices[i] - minPrice;
            }
        }
        return maxProfit;
    }
}

參考

LeetCode