Leetcode121-Best Time to Buy and Sell Stock I - Easy

I

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 (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Note that you cannot sell a stock before you buy one.

Example 1:

Input: [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.
             Not 7-1 = 6, as selling price needs to be larger than buying price.

Example 2:

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

思路：
one pass 算法（在O(N)的时间内解决问题，solve the problem in one pass）
leetcode解释：

The points of interest are the peaks and valleys in the given graph. We need to find the largest peak following the smallest valley. We can maintain two variables - minprice and maxprofit corresponding to the smallest valley and maximum profit (maximum difference between selling price and minprice) obtained so far respectively.

找一个数组中的最大差值（先最小后（最小后出现的）最大）

注意：

1. for (int price : prices) 这个写法更简洁，至于优点...待发掘
2. minprice 的初值设为Integer.MAX_VALUE；
3. maxprofit 的初值设为 0；

代码：

public int maxProfit(int[] prices) {
        int minprice = Integer.MAX_VALUE;
        int maxprofit = 0;
        for (int price : prices) {
            minprice = Math.min(minprice, price);
            maxprofit = Math.max(maxprofit, price - minprice);
        }
        return maxprofit;
    }