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.
给定一个数组，第i个元素为股票在第i天的价格。如果仅允许一次交易，设计算法找出最大的利润。

算法分析

方法一：

嵌套for循环，找出最大的利润。但是时间复杂度为O（n^2）

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

方法二：

一个for循环做两次判断。先判断是不是最小的价格，如果不是最小的价格判断是不是最大的利润

Java代码

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，是否是最大的利润
                maxProfit = prices[i] - minPrice;
            }
        }
        return maxProfit;
    }
}

参考

LeetCode