Leetcode: Best Time to Buy and Sell Stock Leetcode

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.
思路1：
在坐标系中按顺序把这些元素数值画成折线图会好理解一些。需要找出各个最大值和最小值的组合，最小值必须在最大值前面。然后比较各个组合的差值，最大的差值为结果。
Solution1:

public class Solution { public int maxProfit(int[] prices) { if (prices == null || prices.length == 0 || prices.length == 1) { return 0; }int min = prices[0]; int max = prices[0]; int maxPos = 0; int result = 0; for (int i = 1; i < prices.length; i++) { if (min > prices[i]) { if (i > maxPos) { max = prices[i]; maxPos = i; } min = prices[i]; } if (max < prices[i]) { max = prices[i]; maxPos = i; if (result < max - min) result = max - min; } }return result; } }

【Leetcode: Best Time to Buy and Sell Stock】

思路2：遍历卖出时间，寻找卖出时间之前买入价格为最少的值，并想减，取得到的结果中最大的数。

Solution2:

public int maxProfit(int[] prices) { if (prices == null || prices.length == 0 || prices.length == 1) { return 0; }int result = 0; int min = Integer.MAX_VALUE; for (int price : prices) { min = min > price ? price : min; result = result > (price - min) ? result : price - min; }return result; }