Best Time to Buy and Sell Stock

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.

简单点说，就是给定0~i个数分别表示第1至i + 1天的股票价格，问在哪一天买进，哪一天抛出能获得最大利润。首先想到如果i = 1，那么肯定不买也不抛，p1 = 0, i = 2,两天的价格如果先小后大，那必然是第一天买第二天抛，p2为两天价格差，i = 3的时候，先考虑第三天与前两天的最小数之差，将得到的值与p2相比，若p3大于p2,则最大利润更新为p3，否则p3赋值为p2，以此类推：

public class Solution {   public int maxProfit(int[] prices) {    int i = 0;    int len = prices.length;    if (len == 0 || (len == 1) || (len == 2 && (prices[0] > prices[1]))){    return 0;    }    int min[] = new int[len];    min[0] = prices[0];    int p[] = new int[len];    p[0] = 0;    for (i = 1; i < len; i ++){    min[i] = Math.min(Math.min(prices[i - 1], prices[i]), min[i - 1]);    p[i] = Math.max(prices[i] - min[i - 1], p[i - 1]);    }    return p[i - 1];    }}